Chapter 4: Quantities of Reactants and Productsacademic.udayton.edu/VladimirBenin/123_2004/Problem... · 2013. 12. 16. · reactants must add up to the masses of the products, 92.5665

Chapter 4: Quantities of Reactants and Products 121

Chapter 4: Quantities of Reactants and Products

Review Questions

1. A balanced chemical equation indicates the relative amounts of reactants and products so that the number of atoms of each element in the reactants equals the number of atoms of the same elements in the products.

2. Define the problem: Given the equation for a reaction, find the related moles, molecules, and masses of the reactants and products.

Develop a plan: The stoichiometric coefficients (the numbers in front of the molecules’ formulas) in the balanced equation can be interpreted as the related number of moles or the related number of molecules. The molar mass of each molecule is multiplied by the number of moles of that molecule represented in the equation to find its mass in grams.

Execute the plan: 3 H2(g) + N2(g) 2 NH3(g)

The stoichiometric coefficients are 3 for H2, 1 for N2, and 2 for NH3. (Note: Molar masses need only be precise enough to give two decimal places in the calculated masses, comparable to the precision in the given mass of NH3.)

H2 N2 NH3(g)

3 mol 1 mol 2 mol

3 molecules 1 molecule 2 molecules

3 mol × (2.02g/mol) = 6.06 g 1 mol × (28.02g/mol) = 28.02 g 34.08 g

Check your answers: The conservation of mass says that the masses of the reactants must add up to the masses of the products. 6.06 g + 28.02 g = 34.08 g. This looks right.

3. The absolute quantity of each of the reactants is variable and ultimately controlled by the experimenter. The reactants could be present in any relative quantities, though they only react with each other according to the balanced chemical equation. In most cases, that means that one or more of the reactants will be left over when the reaction stops, since it eventually runs out of at least one of the reactants. If, however, the relative quantities of reactants are identical to the relative quantities required by the balanced chemical equation, then the reactants will all run out simultaneously. This special condition is noted by saying that “the reactants were present in stoichiometric amounts.”

4. Define the problem: Given the equation for a reaction, write all the stoichiometric factors represented by it.

Develop a plan: The stoichiometric factors are ratios describing two related quantities. Take each component of the reaction and relate it one by one to the other three components of the reaction, then put each pair-wise relationship in unit factor form.

Execute the plan: 3 MgO(s) + 2 Fe(s) Fe2O3(s) + 3 Mg(s)

3 mol MgO reacts with 2 mol Fe to form 1 mol Fe2O3 and 3 moles Mg.


3 mol MgO2 mol Fe

,

3 mol MgO1 mol Fe2O3

,

3 mol MgO3 mol Mg

or1 mol MgO1 mol Mg

2 mol Fe reacts with 3 mol MgO to form 1 mol Fe2O3 and 3 moles Mg.

2 mol Fe3 mol MgO

,

2 mol Fe1 mol Fe2O3

,

2 mol Fe3 mol Mg

1 mol Fe2O3 is formed with 3 moles Mg when 2 mol Fe reacts with 3 mol MgO.

1 mol Fe2O33 mol Mg

,

1 mol Fe2O32 mol Fe

,

1 mol Fe2O33 mol MgO

3 moles Mg are formed with 1 mol Fe2O3 when 2 mol Fe reacts with 3 mol MgO.

3 mol Mg1 mol Fe2O3

,

3 mol Mg2 mol Fe

,

3 mol Mg3 mol MgO

or1 mol Mg

1 mol MgO

Check your answers: With four different chemicals in this equation we can have 4 × 4 pairs. Since it is not important to relate a molecule to itself, that reduces the number of different pairs by 4. That means we should have (4 × 4) – 4 = 12 stoichiometric factors, and that’s how many we have.

5. If reactants are combined in stoichiometric quantities, they will both run out at the same time. Therefore, all their atoms and the sum of their masses will be present in the lone product, CO2, according to the law of conservation of mass.

10.0 g + 26.6 g = 36.6 g CO2 is produced.

Check your answer: Though it is not necessary, we could check this by using the molar mass to calculate the moles of C or O2 used. Then use the stoichiometry of the balanced equation (C + O2 CO2) to

find the moles of product. Then find the grams using the molar mass of CO2.

10.0 g C ×

1 mol C12.0107 g C

×1 mol CO2

1 mol C×

44.0095 g CO21 mol CO2

= 36.6 g CO2

6. Define the problem: Given a reaction, determine the conversion factors needed to convert between grams and mole, grams and grams, and moles and grams.

Develop a plan: The molar mass and stoichiometric coefficients can be used for the appropriate unit factors.

Execute the plan: 2 Fe(s) + 3 Cl2(g) 2 FeCl3(s)

To take the grams of Cl2 to moles of Cl2, use the molar mass of Cl2:

g Cl2 ×

1 mol Cl270.906 g Cl2

= mol Cl2

To take the moles of Cl2 to moles FeCl3, use the stoichiometry of the balanced equation:

mol Cl2 ×

2 mol FeCl 33 mol Cl 2

= mol FeCl3


To take the grams of FeCl3 to moles of FeCl3, use the molar mass of FeCl3:

g FeCl3 ×

1 mol FeCl 3162.204 g FeCl3

= mol FeCl3

To take the grams of Cl2 to grams of FeCl3, combine all three factors together:

1 mol Cl 270.904 g Cl2

×2 mol FeCl3

3 mol Cl2×

162.204 g FeCl 31 mol FeCl 3

=324.40 g FeCl 3

212.7 g Cl2

g Cl2 ×

324.408 g FeCl 3212.718 g Cl 2

= g FeCl3

Hence, the scheme looks like this:

g Cl 2

324.408 g FeCl3212.718 g Cl2 g FeCl 3

1 mol Cl 270.906 g Cl 2

1 mol FeCl 3162.204 g FeCl 3

mol Cl2

2 mol FeCl 33 mol Cl 2 mol FeCl 3

Check your answers: Going either way around the circuit, the results end up the same.

7. The limiting reactant is the reactant that runs out first. The reaction’s stoichiometry is based on how many reactant molecules are present, not how heavy they are. There are two different reasons why the low mass reactant may not be the limiting reactant. First, the reactant with the smallest mass may be a lightweight element, such as hydrogen. With a low molar mass, the small given mass might still represent a relatively large number of molecules. Second, if the low mass reactant has a relatively small stoichiometric coefficient, then other reactants must be present in larger proportions. Therefore, there is no guarantee that the reactant present with the smallest mass is going to end up being the limiting reactant.

8. A limiting reactant needs to be a reactant. The products are produced in the reaction; therefore they do not qualify.

9. The limiting reactant determines the theoretical yield. If nothing goes wrong and all the reactants that can react do react, then the quantity of limiting reactant should prescribe the quantity of product that will form. The actual yield will only be related to the limiting reactant in that it represents an upper limit to that quantity.

Stoichiometry

10. Define the problem: Given the equation for the reaction, find the related molecules, atoms, moles, masses of each of the reactants and products, and total masses of the reactants and the products.

Develop a plan: The stoichiometric coefficients (the numbers in front of the molecules’ formulas) in the balanced equation can be interpreted as the related number of molecules or the related number of moles of


molecules. The molar mass of each molecule is multiplied by the number of moles of that molecule represented in the equation to find its mass in grams. Add up the mass of KOH and HCl for the total reactant mass. Add up the mass of KCl and H2O for the total product mass.

Execute the plan: KOH(aq) + HCl(aq) KCl(aq) + 6 H2O(l)

The stoichiometric coefficients are 1 for KOH(aq), 1 for HCl(aq), 1 for KCl(aq) and 1 for H2O(l).

KOH HCl KCl H2O

No. molecules 1 1 1 1

No. atoms 1 K, 1 O and 1 H 1 H and 1 Cl 1 K and 1 Cl 2 H and 1 O

No. moles of molecules

1 1 1 1

Mass 1 mol × (56.1056 g/mol) =

56.1056 g

1 mol × (36.4609 g/mol) =

224.00 g

1 mol × (74.5513 g/mol) =

176.04 g

1 mol × (18.0152 g/mol) =

108.12 g

Total mass of reactants

92.5665 g –

Total mass of products

– 92.5665 g

Check your answers: A properly balanced equation has the same numbers of atoms of each type (1K, 1 O, 1 Cl, and 2 H) in the products and reactants. The law of conservation of mass says that masses of the reactants must add up to the masses of the products, 92.5665 g.

11. Define the problem: Given the equation for the reaction, find the related molecules, atoms, moles, masses of each of the reactants and products, and total masses of the reactants and the products.

Develop a plan: The stoichiometric coefficients (the numbers in front of the molecules’ formulas) in the balanced equation can be interpreted as the related number of molecules or the related number of moles of molecules. The molar mass of each molecule is multiplied by the number of moles of that molecule represented in the equation to find its mass in grams. Add up the mass of C2H6 and O2 for the total reactant

mass. Add up the mass of CO2 and H2O for the total product mass.

Execute the plan: 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

The stoichiometric coefficients are 2 for C2H6, 7 for O2, 4 for CO2, and 6 for H2O.

C2H6 O2 CO2 H2O

No. molecules 2 7 4 6

No. atoms 4 C and 12 H 14 O 4 C and 8 O 12 H and 6 O

No. moles of molecules

2 7 4 6

Mass 2 mol × 7 mol × 4 mol × 6 mol ×


(30.0688 g/mol) = 60.16 g

(31.9988 g/mol) = 224.00 g

(44.0095 g/mol) = 176.04 g

(18.0152 g/mol) = 108.12 g

Total mass of reactants

284.16 g –

Total mass of products

– 284.16 g

Check your answers: A properly balanced equation has the same numbers of atoms of each type (4 C, 12 H, and 14 O) in the products and reactants. The law of conservation of mass says that masses of the reactants must add up to the masses of the products, 284.16 g. This looks right.

12. Define the problem: Given the balanced equation for a reaction, identify the stoichiometric coefficients in this equation, and relate the quantity of products to reactants and vice versa.

Develop a plan: (a) The law of conservation of mass says that the total mass of the reactants is the same as the total mass of the products. (b) The stoichiometric coefficients are the numbers in front of each formula in the equation. (c) The stoichiometric coefficients can be interpreted as the related number of reactants. Use the formula stoichiometry of O2 to find the number of molecules that reacted. Use equation stoichiometry to

find out how many atoms of Mg reacted with that many O2 molecules.

Execute the plan: First, balance the equation: 2 Mg(s) + O2(g) 2 MgO(g)

(a) The total mass of product, MgO, is 1.00 g. The total mass of the reactants (mass of Mg plus mass of O2) that reacted must also be 1.00 g, due to the conservation of mass.

(b) The stoichiometric coefficients are 2 for Mg, 1 for O2, and 2 for MgO.

(c) 50 atoms of oxygen make up 25 molecules of O2, since there are two O atoms in each molecule. Since the stoichiometry is 1:1, that means 50 atoms of Mg were needed to react with this much oxygen.

Check your answers: Though it is an unnecessary calculation, we can check the answer to part (a) by calculating the masses of Mg and O2 that reacted to make 1.00 gram of MgO,

1.00 g MgO ×

1 mol MgO40.3044 g MgO

×1 mol Mg

1 mol MgO×

24.3050 g Mg1 mol Mg

= 0.603 g Mg

1.00 g MgO ×

1 mol MgO40.3044 g MgO

×1 mol O2

2mol MgO×

31.9988 g O21 mol O2

= 0.397 g O2

then adding them up: 0.603 g + 0.397 g = 1.000 g. This answer looks right.

13. Define the problem: Given the balanced equation for a reaction and the mass of one reactant, determine how may grams of a product will be formed and how many grams of the other reactant will be required.

Develop a plan: The stoichiometric coefficients can be interpreted as the related number of moles of reactants and products. Use the molar masses to convert mass to moles and the equation stoichiometry to relate moles.

Execute the plan:

1.0 g C12H22O11 ×

1 mol C12H22O11342.2956 g C12H22O11

×12 mol CO2

1 mol C12H22O11×

44.0095 g CO21 mol CO2

= 1.5 g CO2


1.0 g C12H22O11 ×

1 mol C12H22O11342.2956 g C12H22O11

×12 mol O2

1 mol C12H22O11×

31.9988 g O21 mol O2

= 1.1 g O2

Check your answers: The sum of the reactants masses should equal the sum of the product masses:

1.0 g C12H22O11 ×

1 mol C12H22O11342.2956 g C12H22O11

×11 mol H2O

1 mol C12H22O11×

18.0152 g H2O1 mol H2O

= 0.58 g H2O

Adding them: 1.0 g + 1.1 g = 2.1 g, and 1.5 g + 0.58 g = 2.1 g.


14. Define the problem: Write a balanced equation for a combination reaction.

Develop a plan: Use appropriate numbers in front of each reactant and product (stoichiometric coefficients) to balance the equation.

Execute the plan: 4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Check your answer: 4 Fe & 6 O

15. In the first box, three A2 molecules are combined with six B atoms. After the reaction occurs, the box contains only six AB molecules. That means symbolically:

3 A2 + 6 B 6 AB

Therefore, the equation given in (b), A2 + 2 B 2 AB, shows the right stoichiometric relationship between reactants and products.

16. 2 × 4 A2 + 2 × 3 B 2 × 1 B3A8

8 A2 + 6 B 2 B3A8

17. Balance Cl atoms (put a “3” in front of Cl2 and a “2” in front of SbCl3, so each side of the equation has 6 Cl atoms, 3 × 2 = 2 × 3 = 6). Balance Sb atoms (put a 2 in front of Sb, so each side of the equation has 2 Sb atoms): 2 Sb + 3 Cl2 2 SbCl3.

Box (a) represents the correct reactants. Box (c) represents the correct products.

Classification of Chemical Reactions

18. Define the problem: Write the balanced formation equations.

Develop a plan: Write the formula of the compound as the product, and write the reactants in the form of elements. Note, some elements are in the form of diatomic molecules (e.g., O2). Balance the equations.

Execute the plan:

(a) Carbon monoxide = CO(g). 2 C(s) + O2(g) 2 CO(g)

(b) Nickel(II) oxide = NiO(s). 2 Ni(s) + O2(g) 2 NiO(s)

(c) Chromium(III) oxide = Cr2O3(s). 4 Cr(s) + 3 O2(g) 2 Cr2O3(s)

Check your answers: (a) 2 C & 2 O (b) 2 Ni & 2 O (c) 4 Cr & 6 O


19. Define the problem: Write the balanced formation equations.

Develop a plan: Write the formula of the compound as the product, and write the reactants in the form of elements. Note, some elements are in the form of diatomic molecules (e.g., O2). Balance the equations.

Execute the plan:

(a) Copper(I) oxide = Cu2O(s). 4 Cu(s) + O2(g) 2 Cu2O(s)

(b) Arsenic(III) oxide = As2O3(s). 4 As(s) + 3 O2(g) 2 As2O3(s)

(c) Zinc oxide = ZnO(s). 2 Zn(s) + O2(g) 2 ZnO(s)

Check your answers: (a) 4 Cu & 2 O (b) 4 As & 6 O (c) 2 Zn & 2 O

20. Define the problem: Write balanced decomposition equations for some carbonate salts.

Develop a plan: Assume that the products of decomposition are oxides and carbon dioxide as described in the text. Write the formula of the compound as the reactant and determine the formula of the metal oxide. Write the products in the form of the metal oxide and carbon dioxide. Balance the equations.

Execute the plan:

(a) BeCO3(s) BeO(s) + CO2(g) beryllium oxide, carbon dioxide

(b) NiCO3(s) NiO(s) + CO2(g) nickel(II) oxide, carbon dioxide

(c) Al2(CO3)3(s) Al2O3(s) + 3 CO2(g) aluminum oxide, carbon dioxide

Check your answers: (a) 1 Be, 1 C, & 3 O (b) 1 Ni, 1 C, & 3 O (c) 2 Al, 3 C & 9 O

21. Define the problem: Assume that the products of decomposition are oxides and carbon dioxide as described in the text. Write the formula of the compound as the reactant and determine the formula of the metal oxide. Write the products in the form of the metal oxide and carbon dioxide. Balance the equations.

Execute the plan:

(a) ZnCO3(s) ZnO(s) + CO2(g)

zinc oxide, carbon dioxide

(b) MnCO3(s) MnO(s) + CO2(g)

manganese(II) oxide, carbon dioxide

(c) PbCO3(s) PbO(s) + CO2(g)

lead(II) oxide, carbon dioxide

Check your answers: (a) 1 Zn, 1 C, & 3 O (b) 1 Mn, 1 C, & 3 O (c) 1 Pb, 1 C, & 3 O

22. Define the problem: Write the balanced combustion equations.

Develop a plan: The products of combustion depend on the elements present in the compound being combusted:

Compound Contains: C H

Combustion Product is: CO2 H2O


The compound and O2 are the reactant. The products are CO2 and H2O for these combustion reactions.

Balance the equations, starting with C, then H, then O. If the coefficient of O2 ends up a fraction, multiply everything in the reaction by two, to make all coefficients whole numbers.

Execute the plan:

(a) Unbalanced: ? C4H10(g) + ? O2(g) ? CO2(g) + ? H2O(g)

Balance C: C4H10(g) + ? O2(g) 4 CO2(g) + ? H2O(g) 4 C’s

Balance H: C4H10(g) + ? O2(g) 4 CO2(g) + 5 H2O(g) 10 H’s

Balance O: C4H10(g) + 132

O2(g) 4 CO2(g) + 5 H2O(g) 13 O’s

Multiply by 2: 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(g)

(b) Unbalanced: ? C6H12O6(s) + ? O2(g) ? CO2(g) + ? H2O(g)

Balance C: C6H12O6(s) + ? O2(g) 6 CO2(g) + ? H2O(g) 6 C’s

Balance H: C6H12O6(s) + ? O2(g) 6 CO2(g) + 6 H2O(g) 12 H’s

Balance O: C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g) 18 O’s

(c) Unbalanced: ? C4H8O(l) + ? O2(g) ? CO2(g) + ? H2O(g)

Balance C: C4H8O(l) + ? O2(g) 4 CO2(g) + ? H2O(g) 4 C’s

Balance H: C4H8O(l) + ? O2(g) 4 CO2(g) + 4 H2O(g) 8 H’s

Balance O: C4H8O(l) + 112

O2(g) 4 CO2(g) + 4 H2O(g) 12 O’s

Multiply by 2: 2 C4H8O(l) + 11 O2(g) 8 CO2(g) + 8 H2O(g)

Check your answers: (a) 8 C, 20 H, & 26 O (b) 6 C, 12 H, & 18 O (c) 8 C, 16 H, & 24 O

23. Define the problem: Balance the equation for the reaction between a metal and oxygen.

Develop a plan: The products are metal oxides. Write the product in the form of a neutral metal oxide using the periodic table to predict the metal ion’s charge. Balance the equations. Note that the oxide anion has a –2 charge.

Execute the plan: (a) Magnesium is in Group 2A. Its cation will have +2 charge.

2 Mg(s) + O2(g) 2 MgO(s) magnesium oxide

(b) Calcium is in Group 2A. Its cation will have +2 charge.

2 Ca(s) + O2(g) 2 CaO(s) calcium oxide

(c) Indium is in group 3A. Its cation will have +3 charge.

4 In(s) + 3 O2(g) 2 In2O3(s) indium(III) oxide

Check your answers: (a) 2 Mg & 2 O (b) 2 Ca & 2 O (c) 4 In & 6 O


24. Define the problem: Balance the equation for the reaction between a metal and oxygen.

Develop a plan: Write the product formula in the form of a neutral oxide. Note that the Roman numeral in a metal oxide’s name is used to get the metal ion’s charge. Balance the equations. Note that the oxide anion has a –2 charge.

Execute the plan:

(a) Titanium(IV) oxide, contains the titanium(IV) cation with a +4 charge: TiO2

Ti(s) + O2(g) TiO2(s)

(b) Sulfur dioxide is SO2. S8(s) + 8 O2(g) 8 SO2(g)

(c) Selenium dioxide is SeO2. Se(s) + O2(g) SeO2(s)

Check your answers: (a) 1 Ti & 2 O (b) 8 S & 16 O (c) 1 Se & 2 O

25. Define the problem: Balance the equation for the reaction between a metal and a halogen.

Develop a plan: Write the product formula in the form of a neutral halide salt. Balance the equations. Note that the halogens form anions with a –1 charge.

Execute the plan:

(a) Potassium is in group 1A. Its cation will have +1 charge.

2 K(s) + Cl2(g) 2 KCl(s) potassium chloride

(b) Magnesium is in group 2A. Its cation will have +2 charge.

Mg(s) + Br2(l) MgBr2(s) magnesium bromide

(c) Aluminum is in group 3A. Its cation will have +3 charge.

2 Al(s) + 3 F2(g) 2 AlF3(s) aluminum fluoride

Check your answers: (a) 2 K & 2 Cl (b) 1 Mg & 2 Br (c) 2 Al & 6 F

26. Define the problem: Balance the equation for reaction between a metal and a halogen.

Develop a plan: Write the product formula in the form of a neutral halide salt. Note that the Roman numeral in a metal oxide’s name is used to get the metal ion’s charge. Balance the equations. Note that the halogens form anions with a –1 charge.

Execute the plan:

(a) Chromium(III) has a +3 charge, so chromium(III) chloride is: CrCl3

2 Cr(s) + 3 Cl2(g) 2 CrCl3(s)

(b) Copper(II) has a +2 charge, so copper(II) bromide is: CuBr2

Cu(s) + Br2(l) CuBr2(s)

(c) Platinum(IV) has a +4 charge, so platinum(IV) fluoride is: PtF4

Pt(s) + 2 F2(g) PtF4(s)


Check your answers: (a) 2 Cr & 6 Cl (b) 1 Cu & 2 Br (c) 1 Pt & 4 F

Balancing Equations

27. Define the problem: Balance the given equations.

Develop a plan: When balancing equations, select a specific order in which the elements are balanced. Select first the atoms that are only in one product and one reactant, since they will be easier. Select next those elements that are in more than one reactant or product (such as H or O) and then those which are present in elemental form in either the reactants or products. Be systematic. If any of the coefficients end up fractional, multiply every coefficient by the same constant to eliminate the fraction.

Execute the plan:

(a) ? Al(s) + ? O2(g) ? Al2O3(s) Select order: Al then O

2 Al(s) + ? O2(g) 1 Al2O3(s) 2 Al

2 Al(s) + 32

O2(g) 1 Al2O3(s) 3 O

4 Al(s) + 3 O2(g) 2 Al2O3(s)

(b) ? N2(g) + ? H2(g) ? NH3(g) Select order: H then N

? N2(g) + 3 H2(g) 2 NH3(g) 6 H

1 N2(g) + 3 H2(g) 2 NH3(g) 2 N

(c) ? C6H6(l) + ? O2(g) ? H2O(l) + ? CO2(g) Select order: C, H then O

C6H6(l) + ? O2(g) ? H2O(l) + 6 CO2(g) 6 C

C6H6(l) + ? O2(g) 3 H2O(l) + 6 CO2(g) 6 H

C6H6(l) + 152

O2(g) 3 H2O(l) + 6 CO2(g) 15 O

2 C6H6(l) +15 O2(g) 6 H2O(l) + 12 CO2(g)

Check your answers: (a) 4 Al & 6 O (b) 2 N & 6 H (c) 12 C, 12 H & 30 O


Develop a plan: When balancing equations, select a specific order in which the elements are balanced. Select first the atoms that are only in one product and one reactant, since they will be easier. Select next those elements that are in more than one reactant or product (such as H or O) and then those which are present in elemental form in either the reactants or products. Be systematic. If any of the coefficients end up fractional, multiply every coefficient by the same constant to eliminate the fraction.

Execute the plan:

(a) ? Fe(s) + ? Cl2(g) ? FeCl3(s) Select order: Cl then Fe

? Fe(s) + 3 Cl2(g) 2 FeCl3(s) 6 Cl


2 Fe(s) + 3 Cl2(g) 2 FeCl3(s) 2 Fe


(b) ? SiO2(s) + ? C(s) ? Si(s) + ? CO(g) Select order: O, Si then C

SiO2(s) + ? C(s) ? Si(s) + 2 CO(g) 2 O

SiO2(s) + ? C(s) Si(s) + 2 CO(g) 1 Si

SiO2(s) + 2 C(s) Si(s) + 2 CO(g) 2 C

(c) ? Fe(s) + ? H2O(g) ? Fe3O4(s) + ? H2(g) Select order: O, Fe then H

? Fe(s) + 4 H2O(g) Fe3O4(s) + ? H2(g) 4 O

3 Fe(s) + 4 H2O(g) Fe3O4(s) + ? H2(g) 3 Fe

3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g) 8 H

Check your answers: (a) 2 Fe & 6 Cl (b) 1 Si, 2 O & 2 C (c) 3 Fe, 8 H & 4 O


Develop a plan: Follow the method described in the answers to Question 28.

Execute the plan:

(a) ? UO2(s) + ? HF(l) ? UF4(s)+ ? H2O(l) Select order: U, F, H, O

UO2(s) + ? HF(l) UF4(s)+ ? H2O(l) 1 U

UO2(s) + 4 HF(l) UF4(s)+ ? H2O(l) 4 F

UO2(s) + 4 HF(l) UF4(s)+ 2 H2O(l) 4 H and 2 O

(b) ? B2O3(s) + ? HF(l) ? BF3(s) + ? H2O(l) Select order: B, F, H, O

B2O3(s) + ? HF(l) 2 BF3(s) + ? H2O(l) 2 B

B2O3(s) + 6 HF(l) 2 BF3(s) + ? H2O(l) 6 F

B2O3(s) + 6 HF(l) 2 BF3(s) + 3 H2O(l) 6 H and 3 O

(c) ? BF3(g) + ? H2O(l) ? HF(g) + ? H3BO3(s) Select order: B, F, H, O

BF3(g) + ? H2O(l) ? HF(l) + H3BO3(s) 1 B

BF3(g) + ? H2O(l) 3 HF(l) + H3BO3(s) 3 F

BF3(g) + 3 H2O(l) 3 HF(l) + H3BO3(s) 6 H and 3 O

Check your answers: (a) 1 U, 2 O, 4 H & 4 F (b) 2 B, 3 O, 6 H & 6 F (c) 1 B, 3 F, 6 H & 3 O



Execute the plan:

(a) ? MgO(s) + ? Fe(s) ? Fe2O3(s) + ? Mg(s) Select order: O, Mg, Fe


3 MgO(s) + ? Fe(s) Fe2O3(s) + ? Mg(s) 3 O

3 MgO(s) + ? Fe(s) Fe2O3(s) + 3 Mg(s) 3 Mg

3 MgO(s) + 2 Fe(s) Fe2O3(s) + 3 Mg(s) 2 Fe

(b) ? H3BO3(s) ? B2O3(s) + ? H2O(l) Select order: B, H, O

2 H3BO3(s) B2O3(s) + ? H2O(l) 2 B

2 H3BO3(s) B2O3(s) + 3 H2O(l) H and 6 O

(c) ? NaNO3(s) + ? H2SO4(aq) ? Na2SO4(aq) + ? HNO3(g) Select order: Na+, NO3–, H+, SO4

2–

2 NaNO3(s) + ? H2SO4(aq) Na2SO4(aq) + ? HNO3(g) 2 Na+

2 NaNO3(s) + ? H2SO4(aq) Na2SO4(aq) + 2 HNO3(g) 2 NO3–

2 NaNO3(s) + H2SO4(aq) Na2SO4(aq) + 2 HNO3(g) 2 H+ and 1 SO42–

Check your answers: (a) 3 Mg, 3 O & 2 Fe, (b) 6 H, 2 B & 6 O, (c) 2 Na, 2 N, 10 O, 2 H & 1 S



Execute the plan:

(a) ? H2NCl(aq) + ? NH3(g) ? NH4Cl(aq) + ? N2H4(aq) Select order: Cl, N, H

H2NCl(aq) + ? NH3(g) NH4Cl(aq) + ? N2H4(aq) 1 Cl

H2NCl(aq) + 2 NH3(g) NH4Cl(aq) + N2H4(aq) 3 N and 8 H

(b) ? (CH3)2N2H2(l) + ? N2O4(g) ? N2(g) + ? H2O(l) + ? CO2(g) Select order: H, C, O, N

(CH3)2N2H2(l) + ? N2O4(g) ? N2(g) + 4 H2O(g) + ? CO2(g) 8 H

(CH3)2N2H2(l) + ? N2O4(g) ? N2(g) + 4 H2O(g) + 2 CO2(g) 2 C

(CH3)2N2H2(l) + 2 N2O4(g) ? N2(g) + 4 H2O(g) + 2 CO2(g) 8 O

(CH3)2N2H2(l) + 2 N2O4(g) 3 N2(g) + 4 H2O(g) + 2 CO2(g) 6 N

(c) ? CaC2(s) + ? H2O(l) ? Ca(OH)2(s) + ? C2H2(g) Select order: Ca, C, O, H

CaC2(s) + ? H2O(l) Ca(OH)2(s) + ? C2H2(g) 1 Ca

CaC2(s) + ? H2O(l) Ca(OH)2(s) + C2H2(g) 2 C

CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2(g) 2 O and 4 H

Check your answers: (a) 8 H, 3 N & 1 Cl (b) 2 C, 8 H, 6 N & 8 O (c) 1 Ca, 2 C, 4 H & 2 O


32. Define the problem: Balance the given equations. Develop a plan: Follow the method described in the answers to Question 29. Execute the plan:

(a) ? CaNCN(s) + ? H2O(l) ? CaCO3(s) + ? NH3(g) Select order: Ca, N, C, O, H

CaNCN(s) + ? H2O(l) CaCO3(s) + ? NH3(g) 1 Ca

CaNCN(s) + ? H2O(l) CaCO3(s) + 2 NH3(g) 2 N and 1 C

CaNCN(s) + 3 H2O(l) CaCO3(s) + 2 NH3(g) 3 O and 6 H

(b) ? NaBH4(s) + ? H2SO4(aq) ? B2H6(g) + ? H2(g) + ? Na2SO4(g)

Select order: B, Na, S, O, H

2 NaBH4(s) + ? H2SO4(aq) B2H6(g) + ? H2(g) + ? Na2SO4(aq) 2 B

2 NaBH4(s) + ? H2SO4(aq) B2H6(g) + ? H2(g) + Na2SO4(aq) 2 Na

2 NaBH4(s) + H2SO4(aq) B2H6(g) + ? H2(g) + Na2SO4(aq) 1 S and 4 O

2 NaBH4(s) + H2SO4(aq) B2H6(g) + 2 H2(g) + Na2SO4(aq) 10 H

(c) ? H2S(aq) + ? Cl2(aq) ? S8(s) + ? HCl(aq) Select order: S, H, Cl

8 H2S(aq) + ? Cl2(aq) S8(s) + ? HCl(aq) 8 S

8 H2S(aq) + ? Cl2(aq) S8(s) + 16 HCl(aq) 16 H

8 H2S(aq) + 8 Cl2(aq) S8(s) + 16 HCl(aq) 16 Cl

Check your answers: (a) 1 Ca, 2 N, 6 H & 3 O, (b) 2 Na, 2 B, 10 H, 1 S & 4 O

(c) 16 H, 8 S & 16 Cl

33. Define the problem: Balance the given combustion equations.

Develop a plan: When balancing combustion equations, select a C, H, O as the order in which the elements are balanced. Be systematic. If the coefficient of O2 ends up a fraction, multiply every coefficient by 2 to make all of the coefficients whole numbers.

Execute the plan:

(a) ? C6H12O6 + ? O2 ? CO2 + ? H2O

C6H12O6 + ? O2 6 CO2 + ? H2O 6 C

C6H12O6 + ? O2 6 CO2 + 6 H2O 12 H

C6H12O6 + 6 O2 6 CO2 + 6 H2O 18 O

(b) ? C5H12 + ? O2 ? CO2 + ? H2O

C5H12 + ? O2 5 CO2 + ? H2O 5 C

C5H12 + ? O2 5 CO2 + 6 H2O 12 H


C5H12 + 8 O2 5 CO2 + 6 H2O 16 O

(c) ? C7H14 + ? O2 ? CO2 + ? H2O

C7H14O2 + ? O2 7 CO2 + ? H2O 7 C

C7H14O2 + ? O2 7 CO2 + 7 H2O 14 H

C7H14O2 + 192

O2 7 CO2 + 7 H2O 21 O

2 C7H14O2 +19 O2 14 CO2 + 14 H2O

(d) ? C2H4O2 + ? O2 ? CO2 + ? H2O

C2H4O2 + ? O2 2 CO2 + ? H2O 2 C

C2H4O2 + ? O2 2 CO2 + 2 H2O 4 H

C2H4O2 + 2 O2 2 CO2 + 2 H2O 6 O

Check your answers: (a) 6 C, 12 H & 18 O, (b) 5 C, 12 H & 16 O

(c) 14 C, 28 H, 42 O (d) 2 C, 4 H, 6 O



Execute the plan:

(a) ? Mg + ? HNO3 ? H2 + ? Mg(NO3)2 Select order: Mg, N, O, H

Mg + ? HNO3 ? H2 + Mg(NO3)2 1 Mg

Mg + 2 HNO3 ? H2 + Mg(NO3)2 2 N and 6 O

Mg + 2 HNO3 H2 + Mg(NO3)2 2 H

(b) ? Al + ? Fe2O3 ? Al2O3 + ? Fe Select order: Fe, Al, O

? Al + Fe2O3 ? Al2O3 + 2 Fe 2 Fe

2 Al + Fe2O3 Al2O3 + 2 Fe 2 Al and 3 O

(c) ? S + ? O2 ? SO3 Select order: O, S

? S + 3 O2 2 SO3 6 O

2 S + 3 O2 2 SO3 2 S

(d) ? SO3 + ? H2O ? H2SO4 Select order: S, H, O

SO3 + ? H2O H2SO4 1 S

SO3 + H2O H2SO4 2 H and 4 O


Check your answers: (a) 1 Mg, 2 H, 2 N & 6 O, (b) 2 Al, 2 Fe & 3 O

(c) 2 S & 6 O, (d) 1 S, 4 O & 2 H

The Mole and Chemical Reactions

35. Define the problem: Given a balanced chemical equation and the number of moles of a product formed, determine the moles of reactant that was needed.

Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles of product to moles of reactant.

Execute the plan: The balanced equation says: 4 mol HCl are needed to make 1 mol Cl2.

12.5 mol Cl 2 ×

4 mol HCl1 mol Cl 2

= 50.0 mol HCl

Check your answer: More HCl is needed than the Cl2 is formed. It makes sense that the moles of HCl is greater.

36. Define the problem: Given a balanced chemical equation and the number of moles of a reactant, determine the moles of another reactant needed.

Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to moles of the other reactant.

Execute the plan: The balanced equation says: 1 mol CH4 reacts with 2 mol O2.

16.5 mol CH4 ×

2 mol O21 mol CH4

=33.0 mol O2 needed

Check your answer: More O2 is needed than the CH4. It makes sense that the moles of O2 is greater.

37. Define the problem: Given a balanced chemical equation and the number of moles of a reactant, determine the moles and grams of another reactant needed and the grams of product produced.

Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to moles of the other reactant. Use the molar mass to convert from moles to grams. Use the stoichiometry of the equation to find the moles of product formed, then use the molar mass of the product to find the grams.

Execute the plan: The balanced equation says: 2 mol NO react with 1 mol O2.

2.2 mol NO×

1 mol O22 mol NO

= 1.1 mol O2

1.1 mol O2 ×

31.9988 g O21 mol O2

= 35 g O2

The balanced equation says: 2 mol NO produces 2 mol NO2.

2.2 mol NO×

2 mol NO22 mol NO

×46.0055 g NO2

1 mol NO2=1.0 ×102 g NO2 produced


Check your answers: Fewer moles of O2 are needed than moles of NO, so that answer makes sense. The mass of NO used is 2.2 mol × (30.0061 g/mol) = 66 g. The sum of the reactant masses (66 g + 35 g) is equal to the products mass (1.0 × 102 g), within known significant figures. These numbers make sense.

38. Define the problem: Given a balanced chemical equation and the number of moles of a reactant, determine the moles and grams of another reactant needed and the grams of product produced.

Develop a plan: Use the stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to moles of the other reactant. Use the molar mass to convert from moles to grams. Use the stoichiometry of the equation to find the moles of product formed, then use the molar mass of the product to find the grams.

Execute the plan: The balanced equation says: 4 mol Al react with 3 mol O2.

6.0 mol Al ×

3 mol O24 mol Al

= 4.5 mol O2

4.5 mol O2 ×

31.9988 g O21 mol O2

=1.4×102 g O2

The balanced equation says: 4 mol Al produces 2 mol Al2O3.

6.0 mol Al ×

1 mol Al2O32 mol Al

×101.9612 g Al2O3

1 mol Al2O3= 3.1×10 2 g Al2O3produced

Check your answers: More moles of O2 are needed than moles of Al, so that answer makes sense. The

mass of Al used is 6.0 mol × (26.9815 g/mol) = 1.6 × 102 g. The sum of the reactant masses (1.6 × 102 g + 1.4 × 102 g) are equal to the products mass (3.1 × 102 g), within the uncertainty of the significant figures. These numbers make sense.

39. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine the grams of another reactant needed and the moles and grams of product produced.

Develop a plan: Use the molar mass of the first reactant to find the moles of that substance. Then use the stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to moles of the other reactant. Use the molar mass to convert from moles to grams. Use the stoichiometry of the equation to find the moles of product formed, then use the molar mass of the product to find the grams.

Execute the plan: The balanced equation says: 1 mol Fe reacts with 1 mol Cl2.

10 .0 g Fe ×

1 mol Fe55.845 g Fe

= 0.179 mol Fe

0.179 mol Fe ×

1 mol Cl21 mol Fe

×70.906 g Cl 2

1 mol Cl 2=12.7 g Cl 2

The balanced equation says: 1 mol Fe produces 1 mol FeCl2.

0.179 mol Fe×

1mol FeCl21mol Fe

= 0.179 mol FeCl2 expected


0.179 mol FeCl 2 ×

126.751 g FeCl21 mol FeCl2

=22.7 g FeCl 2 expected

Check your answers: The sum of the masses of the reactants must add up to the total mass of the product. 10.0 g + 12.7 g = 22.7 g. This is the same product mass as that calculated above. These numbers make sense.

40. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine the grams of another reactant needed and the moles and grams of product produced.

Develop a plan: Use the molar mass of the first reactant to find the moles of that substance. Then use the stoichiometry of the balanced equation as a conversion factor to convert the moles of the one reactant to moles of the other reactant. Use the molar mass to convert from moles to grams. Use the stoichiometry of the equation to find the moles of product formed, then use the molar mass of the product to find the grams.

Execute the plan:

(a) The balanced equation says: 2 mol Mn react with 3 mol F2.

5.12 g Mn×

1 mol Mn54.9380 g Mn

= 0.0932 mol Mn

0.0932 mol Mn ×

3 mol F22 mol Mn

×37.9996 g F2

1 mol F2= 5.31 g F2

(b) The balanced equation says: 2 mol Mn produces 2 mol MnF3.

0.0932 mol Mn ×

1 mol F21 mol Mn

= 0.0932 mol MnF3 expected

0.0932 mol MnF3 ×

111.9332 g MnF31 mol MnF3

=10.4 g MnF3 expected

Check your answers: The sum of the masses of the reactants must add up to the total mass of the product. 5.12 g + 5.31 g = 10.43 g. This is the same mass (within three significant figures) as that calculated in part (b). These numbers make sense.

41. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine the moles of the reactant and the mass and moles of two products.

Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the balanced equation as a conversion factor to convert the moles of reactant to moles of each product formed, then use the molar mass of each product to find the grams.

Execute the plan: Molar mass of (NH4)2PtCl6 =

2 × [1 × (14.0067 g) + 4 × (1.0079 g)] + (195.078 g) + 6 × (35.453 g) = 443.873 g/mol

12.35 g (NH 4)2 PtCl6 ×

1 mol (NH 4)2 PtCl6443.873 g (NH 4)2 PtCl6

=2.782 ×10−2 mol (NH 4)2 PtCl6

The balanced equation says: 3 mol (NH4)2PtCl6 react to form 3 mol Pt.


2.782 × 10−2 mol (NH 4)2 PtCl6 ×

1 mol Pt1 mol ( NH4)2 PtCl6

=2.782 ×10−2 mol Pt

2.782 × 10−2 mol Pt ×

195.078 g Pt1 mol Pt

= 5.428 g Pt

The balanced equation says: 3 mol (NH4)2PtCl6 react to form 16 mol HCl.

2.782 × 10−2 mol (NH 4)2 PtCl6 ×

16 mol HCl3 mol (NH4 )2 PtCl6

= 0.1484 mol HCl

0.1484 mol HCl ×

36.461 mol HCl1 mol HCl

= 5.410 g HCl

The complete table looks like this:

(NH4)2PtCl6 Pt HCl

12.35 g 5.428 g 5.410 g

0.02782 mol 0.02782 mol 0.1484 mol

Check your answers: The moles are smaller than the grams. This is appropriate. The moles of HCl are larger than the moles of (NH4)2PtCl6 and Pt. These numbers make sense.

42. Define the problem: Given a balanced chemical equation and the number of grams of a product, determine the moles of the product and the mass and moles of the two reactants.

Develop a plan: Use the molar mass of the product to find the moles of that substance. Then use the stoichiometry of the balanced equation as a conversion factor to convert the moles of product to moles of each reactant, then use the molar mass of each reactant to find the grams.

Execute the plan: 103.5 g S2Cl2 ×

1 mol S2Cl 2135.036 g S2Cl 2

= 0.7665 mol S2Cl2

The balanced equation says: 4 mol S2Cl2 is produced from 1 mol S8.

0.7665 mol S2Cl 2 ×

1 mol S84 mol S2Cl2

=0.1916 mol S8

0.1916 mol S8 ×

256.520 g S81 mol S8

= 49.15 g S8

The balanced equation says: 4 mol S2Cl2 is produced from 4 mol S8.

0.7665 mol S2Cl 2 ×

1 mol Cl 21 mol S2Cl2

= 0.7665 mol Cl2

0.7665 mol Cl2 ×

70.906 g Cl 21 mol Cl2

= 54.35 g Cl2

The complete table looks like this:

S8 Cl2 S2Cl2


49.16 g 54.35 g 103.5 g

0.1916 mol 0.7665 mol 0.7665 mol

Check your answers: The sum of the masses of the reactants (49.16 g + 54.35 g = 103.51 g) adds up the mass of the product given (103.5 g). The moles of S8 are smaller than the moles of Cl2 and S2Cl2, because of the smaller stoichiometric coefficient. These numbers make sense.

43. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine the moles of another reactant that are needed and the masses of the two products expected.

Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the balanced equation as a conversion factor to convert the moles of one reactant to moles of the other reactant. Use the stoichiometry to determine the moles of each of the products produced, then use the molar mass of each product to find the grams.

Execute the plan:

(a) The balanced equation says: 1 mol TiCl4 reacts with 2 mol H2O.

14.0 g TiCl 4 ×

1 mol TiCl 4189.688 g TiCl 4

=0.0738 mol TiCl 4

0.0738 mol TiCl 4 ×

2 mol H2O1 mol TiCl 4

=0.148 mol H2O

(b) The balanced equation says: 1 mol TiCl4 produces 1 mol TiO2.

0.0738 mol TiCl 4 ×

1 mol TiO21 mol TiCl 4

×79.866 g TiO2

1 mol TiO2= 5.89 g TiO 2

The balanced equation says: 1 mol TiCl4 produces 4 mol HCl.

0.0738 mol TiCl 4 ×

4 mol HCl1 mol TiCl 4

×36.461 g HCl

1 mol HCl=10.8 g HCl

Check your answers: The sum of the masses of the reactants (14.0 g + 2.67 g = 16.7 g) adds up the mass of the product (5.90 g + 10.8 g = 16.7 g). These numbers make sense.

44. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine the moles of two other reactants needed and the mass of one product expected.

Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the balanced equation as a conversion factor to convert the moles of one reactant to moles of the other reactants. Use the stoichiometry to determine the moles of the product produced, then use the molar mass of the product to find the grams.

Execute the plan:

(a) The balanced equation says: 2 mol SO2 reacts with 2 mol CaCO3.

150. g SO2 ×

1 mol SO264.064 g SO2

×1 mol CaCO 3

1 mol SO2=2.34 mol CaCO3


The balanced equation says: 2 mol SO2 reacts with 1 mol O2.

150. g SO2 ×

1 mol SO264.064 g SO2

×1 mol O2

2 mol SO2=1.17 mol O2

(b) The balanced equation says: 2 mol SO2 produces 2 mol CaSO4.

150. g SO2 ×

1 mol SO264.0638 g SO2

×1 mol CaSO4

1 mol SO2×

136.1406 g CaSO41 mol CaSO4

=319 g CaSO4

Check your answers: The number of moles of O2 is half the number of moles of CaCO3. The calculated moles are smaller than the calculated number of grams. These numbers make sense.

45. Define the problem: Given a balanced chemical equation and the number of grams of a reactant, determine the mass of one product formed.

Develop a plan: Convert from kilograms to grams using a standard metric conversion factor. Then use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the balanced equation to determine the moles of the product produced. Then use the molar mass of the product to find the grams.

Execute the plan: The balanced equation says: 1 mol WO3 produces 1 mol W.

1.00 kg WO3 ×

1000 g WO31 kg WO3

×1 mol WO3

231 .84 g WO3×

1 mol W1 mol WO3

×183.84 g W

1 mol W= 793 g W

Check your answer: It makes sense that the mass of W is smaller than the mass of WO3.

46. Define the problem: Given the formula of a compound and its mass, determine the mass of the elements required to make it.

Develop a plan: Use the molar mass of the compound to find the moles of that substance. Then use the stoichiometry of the balanced equation to determine the moles of each element needed. Then use the molar masses of these elements to find the grams.

Execute the plan: The formula gives: 1 mol GaAs is produced from 1 mol Ga and 1 mol As.

1.45 g GaAs ×

1 mol GaAs144.6446 g GaAs

×1 mol Ga

1 mol GaAs×

69.723 g Ga1 mol Ga

= 0.699 g Ga

1.45 g GaAs ×

1 mol GaAs144.6446 g GaAs

×1 mol As

1 mol GaAs×

74.9215 g As1 mol As

= 0.751 g As

Check your answers: The sum of the reactant masses (0.699 g + 0.751 g = 1.450 g) is the same as the total mass of the compound. These numbers make sense.

47. Define the problem: Given a balanced chemical equation and the mass of the reactant, determine the mass of each of the products produced.

Develop a plan: Use the molar mass of the compound to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the products produced. Then use the molar mass of each product to find the grams.

Execute the plan: The balanced equation says: 2 mol NH4NO3 produces 2 mol N2.


1.0 kg NH4NO3 ×

1000 g NH 4NO31 kg NH4 NO3

×1 mol NH4NO3

80.0432 g NH 4NO3×

2 mol N22 mol NH4NO3

×

28.0134 g N21 mol N2

= 3.5 × 102 g N2

The balanced equation says: 2 mol NH4NO3 produces 4 mol H2O.

1.0 kg NH4NO3 ×


×1 mol NH4NO3

80.0432 g NH 4NO3×

4 mol H2O2 mol NH4NO3

×

18.0152 g H2O1 mol H2O

= 4.5 × 102 g H2O

The balanced equation says: 2 mol NH4NO3 produces 1 mol O2.

1.0 kg NH4NO3 ×


×1 mol NH4NO3

80.0432 g NH 4NO3×

1 mol O22 mol NH4NO3

×

31.9988 g O21 mol O2

= 2.0 × 102 g O2

Check your answers: The sum of the product masses

3.5 × 102 g + 4.5 × 102 g + 2.0 × 102 g = 1.00 × 103 g

is the same as the total mass of the reactant compound. These numbers make sense.

48. Define the problem: Given the products and reactants of a reaction and the mass of one reactant, balance the chemical equation, determine the mass of the product produced, and determine the mass of the other reactant.

Develop a plan: Balance the equation from the given formulas of the reactants and product. Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the product produced and the moles of the other reactant required. Then use the molar mass of each to find the grams.

Execute the plan: (a) Balance O atoms, then Fe atoms.

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

(b) The balanced equation says: 4 mol Fe produces 2 mol Fe2O3.

5.58 g Fe ×

1 mol Fe55.845 g Fe

×2 mol Fe2O3

4 mol Fe×

159.688 g Fe 2O31 mol Fe2O3

= 7.98 g Fe2O3

(c) The balanced equation says: 4 mol Fe reacts with 3 mol O2.

5.58 g Fe ×

1 mol Fe55.845 g Fe

×3 mol O24 mol Fe

×31.9988 g O2

1mol O2= 2.40 g O2

Check your answers: The sum of the reactant masses (5.58 g + 2.40 g = 7.98 g)

is the same as the total mass of the product compound. These numbers make sense.

49. Define the problem: Given a balanced chemical equation and the mass of the reactant, determine the masses of all the products produced.


Develop a plan: Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the products produced. Then use the molar mass of each product to find the grams.

Execute the plan:

1.0 g C3H5( NO3)3 ×

1 mol C3H5(NO3)3227.0863 g C3H5(NO3)3

=0.0044 mol C3H5(NO3)3

The balanced equation says: 4 mol C3H5(NO3)3 produces 12 mol CO2.

0.0044 mol C3H5(NO3)3 ×

12 mol CO24 mol C3H5( NO3)3

×44.0095 g CO2

1 mol CO2 = 0.58 g CO2

The balanced equation says: 4 mol C3H5(NO3)3 produces 10 mol H2O.

0.0044 mol C3H5(NO3)3 ×

10 mol H2O4 mol C3H5(NO3)3

×18.0152 g H2O

1 mol H2O = 0.20 g H2O

The balanced equation says: 4 mol C3H5(NO3)3 produces 6 mol N2.

0.0044 mol C3H5(NO3)3 ×

6 mol N24 mol C3H5( NO3)3

×28.0134 g N2

1 mol N2 = 0.19 g N2

The balanced equation says: 4 mol C3H5(NO3)3 produces 1 mol O2.

0.0044 mol C3H5(NO3)3 ×

1 mol O24 mol C3H5(NO3)3

×31.9988 g O2

1 mol O2 = 0.035 g O2

Check your answers: The sum of the product masses (0.58 g + 0.20 g + 0.19 g + 0.035 g = 1.01 g) is the same (to two significant figures) as the total mass of the reactant compound (1.0 g). These numbers make sense.

50. Define the problem: Given the reactants and products of a reaction and the mass of a reactant, balance the chemical equation, determine the mass of another reactant, and determine the mass of one product produced.

Develop a plan: Given the formulas of the reactants and products, balance the equation, selecting appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the other reactant and product produced. Then use the molar mass of each to find the grams.

Execute the plan:

(a) ? CCl2F2 + ? Na2C2O4 ? C + ? CO2 + ? NaCl + ? NaF

Select order: Cl, F, Na, O, C

CCl2F2 + ? Na2C2O4 ? C + ? CO2 + 2 NaCl + ? NaF 2 Cl

CCl2F2 + ? Na2C2O4 ? C + ? CO2 + 2 NaCl + 2 NaF 2 F

CCl2F2 + 2 Na2C2O4 ? C + ? CO2 + 2 NaCl + 2 NaF 4 Na

CCl2F2 + 2 Na2C2O4 ? C + 4 CO2 + 2 NaCl + 2 NaF 8 O


CCl2F2 + 2 Na2C2O4 C + 4 CO2 + 2 NaCl + 2 NaF 5 C

(b) The balanced equation says: 1 mol CCl2F2 requires 2 mol Na2C2O4.

76.8 g CCl 2F2 ×

1 mol CCl 2F2120.914 g CCl 2F2

= 0.635 mol CCl 2F2

0.635 mol CCl 2F2 ×

2 mol Na 2C2O41 mol CCl 2F2

×133.9986 g Na 2C2O4

1 mol Na 2C2O4=170. g Na 2C2O4

The balanced equation says: 1 mol CCl2F2 produces 4 mol CO2.

0.635 mol CCl 2F2 ×

4 mol CO21 mol CCl2F2

×44.0095 g CO2

1 mol CO2=112 g CO2

Check your answers: (a) Check the number of atom of each type in the reactants and products: 5 C, 2 Cl, 2 F, 4 Na, 8 O. The equation is properly balanced. (b) The masses of Na2C2O4 and CO2 are both larger than

the mass of CCl2F2. That makes sense, looking at the stoichiometry and the molar masses.

51. Define the problem: Given the reactants and products of a reaction and the mass of a reactant, balance the chemical equation, determine the mass of another reactant, and determine the mass of one product produced.

Develop a plan: Given the formulas of the reactants and products, balance the equation, selecting appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the other reactant and product produced. Then use the molar mass of each to find the grams.

Execute the plan:

(a) ? NH4NO3 ? N2O + ? H2O Select order: N, H, O

NH4NO3 N2O + ? H2O 2 N

NH4NO3 N2O + 2 H2O 4 H and 3 O

(b) The balanced equation says: 1 mol NH4NO3 requires 1 mol N2O.

10.0 g NH4NO3 ×

1 mol NH4NO380.0432 g NH 4NO3

=0.125 mol NH4NO3

0.125 mol NH4 NO3 ×

1 mol N2O1 mol NH 4NO3

×44.0128 g N2O

1 mol N2O= 5.50 g N2O

The balanced equation says: 1 mol NH4NO3 requires 2 mol H2O.

0.125 mol NH4 NO3 ×

2 mol H2O1 mol NH 4NO3

×18.0152 g H2O

1 mol H2O= 4.50 g H2O

Check your answers: (a) Check the number of atom of each type in the reactants and products: 2 N, 4 H, 3 O. The equation is properly balanced. (b) The sum of the masses of the products (5.50 g + 4.50 g = 10.00 g) is equal to the mass of the reactant compound (10.0 g). These numbers make sense.


52. Define the problem: Given a balanced chemical equation and the mass of a reactant in kilograms, determine the mass of one of the products produced and the mass of another reactant required.

Develop a plan: Use metric conversion factors to convert kilograms to grams. Use the molar mass of the reactant to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the product produced and the other reactant required. Then use molar masses to find the grams.

Execute the plan: 1.00 kg Fe2O3 ×

1000 g Fe2O31 kg Fe2O3

×1 mol Fe2O3

159.688 g Fe2O3= 6.26 mol Fe2O3

(a) The balanced equation says: 1 mol Fe2O3 produces 1 mol Fe.

6.26 mol Fe2O3 ×

2 mol Fe1 mol Fe 2O3

×55.845 g Fe

1 mol Fe= 699 g Fe

(b) The balanced equation says: 1 mol Fe2O3 requires 3 mol CO.

6.26 mol Fe2O3 ×

3 mol CO1 mol Fe 2O3

×28.0101 g CO

1 mol CO= 526 g CO

Check your answers: The mass of iron produced is less than the mass of iron(III) oxide it was produced from. The mass of CO used is less than the mass of iron(III) oxide even with a larger stoichiometric coefficient because CO contains lighter-weight atoms.

53. Define the problem: Given the reactants and products of a reaction and the desired mass of a product, balance the chemical equation and determine the mass of the reactants needed to produce that product.

Develop a plan: Given the formulas of the reactants and products, balance the equation, selecting appropriate order for the systematic balancing of all the atoms. Then use the molar mass of the product to find the moles of that substance. Then use the stoichiometry of the equation to determine the moles of the reactants needed to make that much product. Then use the molar mass of each to find the grams.

Execute the plan:

(a) ? K2PtCl4 + ? NH3 ? Pt(NH3)2Cl2 + ? KCl Select order: N, H, Pt, K, Cl

? K2PtCl4 + 2 NH3 Pt(NH3)2Cl2 + ? KCl 2 N and 6 H

K2PtCl4 + 2 NH3 Pt(NH3)2Cl2 + ? KCl 1 Pt

K2PtCl4 + 2 NH3 Pt(NH3)2Cl2 + 2 KCl 2 K and 4 Cl

(b) The balanced equation says: 1 mol Pt(NH3)2Cl2 requires 1 mol K2PtCl4.

2.50 g Pt(NH3)2Cl 2 ×

1 mol Pt( NH3)2Cl 2300.045 g Pt(NH3)2Cl 2

= 0.00833 mol Pt(NH3)2Cl2

0.00833 mol Pt(NH 3)2 Cl2 ×

1 mol K2PtCl41 mol Pt(NH3)2Cl 2

×415.0866 g K2PtCl4

1 mol K2PtCl4 = 3.46 g K2PtCl4

The balanced equation says: 1 mol Pt(NH3)2Cl2 requires 2 mol NH3.

0.00833 mol Pt(NH 3)2 Cl2 ×

2 mol NH31 mol Pt(NH3)2Cl 2

×17.0304 g NH3

1 mol NH 3 = 0.284 g NH3


Check your answers: (a) Check the number of atom of each type in the reactants and products: 2 K, 1 Pt, 2 N, 6 H, and 4 Cl. The equation is properly balanced. (b) The mass of K2PtCl4 is larger than the mass of

Pt(NH3)2Cl2, and the mass of NH3 is less than the mass of Pt(NH3)2Cl2. These both make sense, looking at the stoichiometric relationships and the molar masses.

Limiting Reactant

54. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the limiting reactant and the mass of the product produced.

Develop a plan: Here, we use a slight variation of what the text calls “the mole method.” We will calculate a directly comparable quantity, the moles of the desired product. Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. Then, use the moles of product produced from the limiting reactant and the molar mass of the product to find the grams.

Execute the plan: Exactly 1 gram of each reactant is present initially.

The balanced equation says: 1 mol Na2SO4 produces 1 mol BaSO4.

1 g Na 2SO4 ×

1 mol Na 2SO4142.042 g Na 2SO 4

×1 mol BaSO4

1 mol Na 2SO4= 0.00704016 mol BaSO4

The balanced equation says: 1 mol BaCl2 produces 1 mol BaSO4.

1 g BaCl 2 ×

1 mol BaCl 2208.233 g BaCl 2

×1 mol BaSO41 mol BaCl 2

= 0.00480231 mol BaSO4

The number of BaSO4 moles produced from BaCl2 is smaller (0.00480231 mol < 0.00704016 mol), so BaCl2 is the limiting reactant.

Find the mass of 0.00480231 mol BaSO4:

0.00480231 mol BaSO4 ×

233.390 g BaSO 41 mol BaSO4

=1.12081 g BaSO4

Check your answers: There are fewer moles of BaSO4 present than BaCl2, and the equation needs the same

amount of BaSO4 as BaCl2, so it makes sense that BaCl2 is the limiting reactant.

55. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the excess reactant and the mass left over when the product produced.

Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the excess reactant from the reactant that produces the most number of products. The reactant that produces the least moles of product is the limiting reactant. From the mass of the limiting reactant, determine the moles of the other reactant needed for complete reaction. Convert that number to grams using the molar mass. Then subtract the quantity used from the initial mass given to get the mass of excess reactant.

Execute the plan: The balanced equation says: 2 mol Al produces 1 mol Al2O3.


100. g Al×

1 mol Al26.9815 g Al

×1 mol Al2O3

2 mol Al= 1.85 mol Al2O3

The balanced equation says: 3 mol MnO produces 1 mol Al2O3.

200. g MnO×

1 mol MnO70.9374 g MnO

×1 mol Al2O33 mol MnO

= 0.940 mol Al2O3

The number of Al2O3 moles produced from Al is larger (1.85 mol > 0.940 mol), so Al is the excess reactant

and MnO is the limiting reactant.

The balanced equation says: 3 mol MnO is produced from 2 mol Al.

200. g MnO×

1 mol MnO70.9374 g MnO

×2 mol Al

3 mol MnO×

26.9815 g Al1 mol Al

= 50.7 g Al used

100. g Al initial – 50.7 g Al used up = 49 g Al remains unreacted

Check your answers: There are fewer moles of MnO present than Al, and the equation needs more MnO than Al, so it makes sense that MnO is the limiting reactant. In addition, the calculation of excess reactant remaining proved that the initial mass of Al was larger than required to react with all of the MnO.

56. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the limiting reactant, the mass of the product produced, and the mass remaining of the excess reactant when the reaction is complete.

Develop a plan: Here, we use a slight variation of what the text calls “the mole method.” We will calculate a directly comparable quantity, the moles of product. (a) Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. (b) Use the moles of product produced from the limiting reactant and the molar mass of the product to find the grams. (c) From the limiting reactant quantity, determine the moles of the other reactant needed for complete reaction. Convert that number to grams using the molar mass. Then subtract the quantity used from the initial mass given to get the mass of excess reactant.

Execute the plan:

(a) The balanced equation says: 2 mol Al produces 2 mol AlCl3.

2.70 g Al×

1 mol Al26.9815 g Al

×2 mol AlCl 3

2 mol Al= 0.100 mol AlCl 3

The balanced equation says: 3 mol Cl2 produces 2 mol AlCl3.

4.05 g Cl2 ×

1 mol Cl 270.906 g Cl 2

×2 mol AlCl 3

3 mol Cl2= 0.0381 mol AlCl 3

The number of AlCl3 moles produced from Cl2 is smaller (0.0381 mol < 0.100 mol), so Cl2 is the limiting reactant and Al is the excess reactant.

(b) Find the mass of 0.0381 mol AlCl3:

0.0381 mol AlCl 3 ×

133.341 g AlCl 31 mol AlCl3

= 5.08 g AlCl 3


(c) The balanced equation says: 3 mol Cl2 react with 2 mol Al.

4.05 g Cl2 ×

1 mol Cl270.906 g Cl2

×2 mol Al

3 mol Cl2×

26.9815 g Al1 mol Al

=1.03 g Al

2.70 g Al initial – 1.03 g Al used up = 1.67 g Al remains unreacted

Check your answers: There are fewer moles of Cl2 present than Al, and the equation needs more Cl2 than

Al, so it makes sense that Cl2 is the limiting reactant. In addition, the calculation in (c) proved that the initial

mass of Al was larger than required to react with all of the Cl2. The sum of the masses of the reactants that reacted (4.05 g + 1.03 g = 5.08 g) equals the mass of the product produced (5.08 g). These answers make sense.

57. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the limiting reactant and the mass of the product produced.

Develop a plan: (a) Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. (b) Use the moles of product produced from the limiting reactant and the molar mass of the product to find the grams.

Execute the plan:

(a) The balanced equation says: 2 mol H2 produces 2 mol H2O.

100. g H2 ×

1 mol H22.0158 g H2

×2 mol H2O2 mol H2

= 49.6 mol H2O

The balanced equation says: 1 mol O2 produces 2 mol H2O.

100. g O2 ×

1 mol O231.9988 g O2

×2 mol H2O1 mol O2

= 6.25 mol H2O

The number of H2O moles produced from O2 is smaller (6.25 mol < 49.6 mol), so O2 is the limiting

reactant and H2 is the excess reactant.

(b) Find the mass of 6.25 mol H2O:

6.25 mol H2O ×

18.0152 g H2O1 mol H2O

=113 g H2O

Check your answers: The reactants are present in the same mass quantities, but the molar mass of H2 is far

smaller than the molar mass of O2, so there are far fewer O2 molecules. Therefore, it makes sense that O2 is the limiting reactant.


Develop a plan: (a) Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. (b) From the limiting reactant quantity determine the moles of the other reactant needed for complete reaction. Convert that


number to mass using the molar mass. Then subtract the quantity used from the initial mass given, to get the mass of excess reactant. (c) Use the moles of product produced from the limiting reactant and the molar mass of the product to find the grams.

Execute the plan: (a) The balanced equation says: 1 mol CO produces 1 mol CH3OH.

74 .5 g CO ×

1 mol CO28.0101 g CO

×1 mol CH3OH

1 mol CO= 2.66 mol CH 3OH

The balanced equation says: 2 mol H2 produces 1 mol CH3OH.

12.0 g H2 ×

1 mol H22.0158 g H2

×1 mol CH3OH

2 mol H2= 2.98 mol CH 3OH

The number of CH3OH moles produced from CO is smaller (2.66 mol < 2.98 mol), so CO is the limiting

reactant and H2 is the excess reactant.

(b) The balanced equation says: 2 mol H2 react with 1 mol CO.

74.5 g CO ×

1 mol CO28.0101 g CO

×2 mol H21 mol CO

×2.0158 g H2

1 mol H2=10.7 g H2

12.0 g H2 initial – 10.7 g H2 used up = 1.3 g H2 remains unreacted

(c) Find the mass of 2.66 mol CH3OH:

2.66 mol CH3OH ×

32.0417 g CH 3OH1 mol CH 3OH

= 85.2 g CH 3OH

Check your answers: The calculation in (b) proved that the initial mass of H2 was larger than required to react with all of the CO. The sum of the masses of the reactants that reacted (74.5 g + 10.7 g = 85.2 g) equal the mass of the product produced (85.2 g). These answers make sense.

59. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the moles of reactants and products present when the reaction is finished.

Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances present initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify the limiting reactant from the reactant that produces the least number of products. From the limiting reactant quantity, and using the stoichiometry of the balanced equation, determine the moles of the other reactant and products.

Execute the plan: The balanced equation says: 1 mol CH4 produces 1 mol CO2.

995 g CH 4 ×

1 mol CH 416.0423 g CH4

×1 mol CO21 mol CH4

=62.0 mol CO2

The balanced equation says: 2 mol H2O produces 1 mol CO2.

2510 g H2O×

1 mol H2O18.0152 g H2O

×1 mol CO22 mol H2O

= 69.6 mol CO2


The number of CO2 moles produced from CH4 is smaller (62.0 mol < 69.6 mol), so CH4 is the limiting reactant

and H2O is the excess reactant.

The balanced equation says: 1 mol CO2 is produced with 4 mol H2.

62.0 mol CO2 ×

4 mol H21 mol CO2

= 248 mol H2

The balanced equation says: 1 mol CH4 reacts with 2 mol H2O.

995 g CH 4 ×

1 mol CH 416.0423 g CH4

×2 mol H2O1 mol CH 4

= 124 mol H2O reacted

2510 g H2O×

1 mol H2O18.0152 g H2O

= 139 mol H2O present initially

139 mol H2O initial – 124 mol H2O reacted = 15 mol H2O left

Check your answers: The excess moles of H2O prove that the right limiting reactant was determined, since

there are no moles of CH4 left. The moles of products are stoichiometrically appropriate multiples of 62.0 moles, as they should be. These numbers look right.

60. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the moles of reactants and products present when the reaction is finished.

Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances present initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify the limiting reactant from the reactant that produces the least number of products. From the limiting reactant quantity, and using the stoichiometry of the balanced equation, determine the moles of the other reactant and products.

Execute the plan: The balanced equation says: 1 mol CaO produces 1 mol H2O.

112 g CaO ×

1 mol CaO56.077 g CaO

×1 mol H2O1 mol CaO

= 2.00 mol H2O

The balanced equation says: 2 mol NH4Cl produces 1 mol H2O.

224 g NH4Cl ×

1 mol NH 4Cl53.4913 g NH 4Cl

×1 mol H2O

2 mol NH4Cl= 2.09 mol H2O

The number of H2O moles produced from CaO is smaller (2.00 mol < 2.09 mol), so CaO is the limiting reactant

and NH4Cl is the excess reactant.

The balanced equation says: 1 mol CaO produces 2 mol NH3.

112 g CaO ×


×2 mol NH31 mol CaO

= 4.00 mol NH3

The balanced equation says: 1 mol CaO produces 1 mol CaCl2.


112 g CaO ×


×1 mol CaCl21 mol CaO

= 2.00 mol CaCl 2

The balanced equation says: 1 mol CaO reacts with 2 mol NH4Cl.

112 g CaO ×


×2 mol NH4Cl

1 mol CaO= 4.00 mol NH4Cl reacted

224 g NH4Cl ×

1 mol NH4Cl53.4913 g NH4Cl

= 4.19 mol NH4Cl present initially

4.19 mol NH4Cl initial – 4.00 mol NH4Cl reacted = 0.19 mol NH4Cl left

Check your answers: The excess moles of NH4Cl prove that the right limiting reactant was determined, since there are zero moles of CaO left. The moles of products are stoichiometrically appropriate multiples of 2.00 moles, as they should be. These numbers look right.

61. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the maximum amount of product that can be produced.

Develop a plan: Use a metric conversion and the molar mass of the reactants to find the moles of the reactant substances present initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify the limiting reactant from the reactant that produces the least number of products. From the limiting reactant quantity, determine the moles of the product, then use the molar mass of the product to get the grams and a metric conversion to get kilograms.

Execute the plan: The balanced equation says: 1 mol Fe2O3 produces 2 mol Fe.

2.00 kg Fe2O3 ×


×1 mol Fe2O3

159.6882 g Fe 2O3×

2 mol Fe1 mol Fe2O3

= 25.0 mol Fe

The balanced equation says: 3 mol CO produces 2 mol Fe.

2.00 kg CO ×

1000 g CO1 kg CO

×1 mol CO

28.0101 g CO×

2 mol Fe3 mol CO

= 47.6 mol Fe

The number of Fe moles produced from Fe2O3 is smaller (25.0 mol < 47.6 mol), so Fe2O3 is the limiting reactant and CO is the excess reactant.

Find the mass of the Fe from the limiting reactant:

25 .0 mol Fe ×

55.845 g Fe1 mol Fe

×1 kg Fe

1000 g Fe= 1.40 kg Fe

Check your answer: The reactants are present in the same mass quantities, but Fe2O3 has a larger molar mass. So, it makes sense that it is the limiting reactant. The mass of iron should be smaller than the mass of iron (III) oxide. These numbers look right.

62. Define the problem: Given a balanced chemical equation and the masses of both reactants, determine the maximum amount of product that can be produced.

Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances present initially. Then use the stoichiometry of the equation to determine the moles of one of the products. Identify


the limiting reactant from the reactant that produces the least number of products. From the limiting reactant quantity, determine the moles of the product, then use the molar mass of the product to get the grams.

Execute the plan: The balanced equation says: 2 mol C7H6O3 produces 2 mol C9H8O4.

100. g C7 H6O3 ×

1 mol C7 H6O3138.1205 g C7 H6O3

×2 mol C9H8O42 mol C7 H6O3

=0.724 mol C9H8O4

The balanced equation says: 1 mol C4H6O3 produces 2 mol C9H8O4.

100. g C4H6O3 ×

1 mol C4H6O3102.0884 g C4H6O3

×2 mol C9H8O41 mol C4H6O3

=1.96 mol C9H8O4

The number of C9H8O4 moles produced from C7H6O3 is smaller (0.724 mol < 1.96 mol), so C7H6O3 is the

limiting reactant and C4H6O3 is the excess reactant.

Find the mass of the C9H8O4 from the moles of limiting reactant:

0.724 mol C9H8O4 ×

180.1571 g C9H8O41 mol C9H8O4

= 130. g C9H8O4

Check your answers: The reactants are present in the same mass quantities, but C7H6O3 has a larger molar mass and needs two moles rather than just one mole to react. So, it makes sense that it is the limiting reactant.

Percent Yield

63. Define the problem: Given the balanced chemical equation, the mass of the limiting reactant and the actual yield, determine the theoretical yield and the percent yield.

Develop a plan: First, calculate the theoretical yield by determining the maximum mass of product that could have been made from the given quantity of reactant: Take the mass of the limiting reactant and convert it to moles. Then use stoichiometry to find the moles of product. Then convert to grams using molar mass. Take the given actual yield and divide by the calculated theoretical yield and multiply by 100 % to get percent yield.

Execute the plan: The limiting reactant is Fe2O3. From its mass, find the maximum grams of Fe that could be made. The mole ratio comes from the balanced equation.

1.00 kg Fe2O3 ×


×1 mol Fe2O3

159.688 g Fe2O3×

2 mol Fe1 mol Fe 2O3

×55.845 g Fe

1 mol Fe = 699 g Fe

The given mass of Fe is the actual yield.

654 g Fe actual699 g Fe theoretical

× 100 % = 93.5 % yield

Check your answer: Close to the maximum quantity of iron was produced, so it makes sense that the percent yield is over ninety percent.

64. Define the problem: Given the theoretical yield and the actual yield, determine the percent yield.


Develop a plan: Divide the actual yield by the theoretical yield and multiply by 100 % to get percent yield. (Note that the balanced equation is given too, but you don’t need to use it to answer this question.)

Execute the plan:

100 g NH 3 actual136 g NH3 theoretical

×100 % = 73.5 % yield

Check your answer: About three-quarters of the maximum quantity of ammonia was produced, so it makes sense that the percent yield is about 75 %. This number looks right.

65. Define the problem: Given the theoretical yield and the actual yield, determine the percent yield.

Develop a plan: Divide the actual yield by the theoretical yield and multiply by 100 % to get percent yield. (Note that the balanced equation is given, too, but you don’t need to use it to answer this question.)

Execute the plan

36.7 g CaO actual65.5 g CaO theoretical

×100 % = 56.0 % yield

Check your answer: A little more than half the maximum quantity of quicklime was produced, so it makes sense that the percent yield is a little more than 50 %. This number looks right.

66. Define the problem: Given the balanced chemical equation, the mass of the limiting reactant and the actual yield, determine the percent yield.

Develop a plan: First, we need to calculate the theoretical yield. We get this by determining the maximum amount of product that could have been made from the given quantities of reactant: Take the mass of the limiting reactant and convert it to moles. Then use stoichiometry to find the moles of product. Then convert to grams using molar mass. Take the given actual yield and divide by the calculated theoretical yield and multiply by 100 % to get percent yield.

Execute the plan: The limiting reactant is NaBH4. From its mass, find the maximum grams of B2H6 that could be made. The mole ratio comes from the balanced equation.

1.203 g NaBH4 ×

1 mol NaBH437.832 g NaBH 4

×1 mol B2H6

2 mol NaBH 4×

27.669 g B2H61 mol B2H6

= 0.4399 g B2H6

The given mass of B2H6 is the actual yield.

0.295 g B2H6 actual0.4399 g B2H6 theoretical

× 100 % = 67.1 % yield

Check your answer: About two-thirds the maximum quantity of diborane was produced, so it makes sense that the percent yield is about 67 %. This number looks right.

67. Define the problem: Given the balanced chemical equation, the mass of the limiting reactant and the actual yield, determine the percent yield.

Develop a plan: First, we need to calculate the theoretical yield. We get this by determining the maximum amount of product that could have been made from the given quantities of reactant: Take the mass of the limiting reactant and convert it to moles. Then use stoichiometry to find the moles of product. Then convert to grams using molar mass. Take the actual yield and divide by the calculated theoretical yield and multiply by 100 % to get percent yield.

Execute the plan: The limiting reactant is H2. From its mass, find the maximum grams of CH3OH that could be made. The mole ratio comes from the balanced equation.


5.0×103 g H2 ×

1 mol H22.0158 g H2

×1 mol CH3OH

2 mol H2×

32.0417 g CH 3OH1 mol CH 3OH

= 4.0 ×10 4 g CH3OH

The given mass of CH3OH is the actual yield.

3.5×10 3 g CH 3OH actual

4.0 ×104 g CH3OH theoretical×100 % =8.8 % yield

Check your answer: The actual mass produced is roughly a factor of ten less than the maximum quantity of ethanol produced, so it makes sense that the percent yield is around 10 %. This number looks right.

68. Define the problem: Given the balanced chemical equation, the desired mass of the product, and the percent yield, determine the amount of limiting reactant that must be used.

Develop a plan: Interpret the percent yield as the relationship between the actual grams and theoretical grams. Use that relationship to find the theoretical yield mass of the product. Then use stoichiometry to find the moles of limiting reactant. Then, using molar mass, convert to actual grams of reactant needed.

Execute the plan: The 51 % percent yield tells us the following:

To make 51 grams of S2Cl2, we need to have enough limiting reactant to make 1.19 grams of S2Cl2.

1.19 g S2Cl 2 actual×

100. g S2Cl 251 g S2Cl2 actual

= 2.3 g S2Cl2

Use this mass of product to determine what mass of limiting reactant to use.

2.3 g S2Cl2 ×

1 mol S2Cl 2135.036 g S2Cl2

×3 mol SCl 2

1 mol S2Cl 2×

102.971 g SCl 21 mol S2Cl 2

= 5.3 g SCl 2

Check your answer: The yield suggests that we need to try to make about twice as much, so the 2.3 grams makes sense. The stoichiometry is 3:1, so it makes sense that a larger mass of SCl2 is needed than the mass

of S2Cl2 formed. These numbers look right.

69. Define the problem: Given the balanced chemical equation, the desired mass of the product, and the percent yield, determine the amount of limiting reactant that must be used.

Develop a plan: Interpret the percent yield as the relationship between the actual grams and theoretical grams. Use that relationship to find the theoretical yield mass of the product. Then use stoichiometry to find the moles of limiting reactant. Then using molar mass, convert to actual grams of reactant needed.

Execute the plan: The 92 % percent yield tells us the following:

To make 92 grams of Si3N4, we need to have enough limiting reactant to make 100. grams of Si3N4.

1.0 kg Si3N4 actual ×

1000 g Si3N41 kg Si3N4

×100. g Si3N4

92 g Si3N4 actual=1.1× 103 g Si3N4

Use this mass of product to determine what mass of limiting reactant to use.

1.1×103 g Si3N4 ×

1 mol Si3N4140.2833 g Si3N4

×3 mol Si

1 mol Si3N4×

28.0855 g Si1 mol Si

= 660 g Si


Check your answer: The yield suggests that we don’t need to try to make a lot more, so the 1.1 kilograms makes sense. The product mass is smaller because Si has a significantly smaller molar mass than Si3N4. These numbers look right.

Empirical Formulas

70. Define the problem: Given the percent mass of elements in a compound, determine the empirical formula.

Develop a plan: Choose a convenient sample mass of SxOy, such as 100.0 g. Find the mass of S and O in

the sample, using the given mass percent. Use the molar mass of the elements to find their moles, then use a whole-number mole ratio to determine the empirical formula.

Execute the plan: 100.0 g of .SxOy contains 40.0 g S and 60.0 g O.

40.0 g S ×

1 mol S32.065 g S

= 1.25 mol S 60 .0 g O ×

1 mol O15.9994 g O

= 3.75 mol O

Set up mole ratio and simplify by dividing by the smallest number of moles:

1.25 mol S : 3.75 mol O

1 S : 3 O

Use the whole number ratio for the subscripts in the formula. The empirical formula is SO3.

Check your answer: The percent by mass of oxygen in the compound is somewhat greater than the percent by mass of sulfur. Since sulfur atoms weigh more than oxygen atoms, it makes sense that there are several O atom per S atom present in this formula. SO3 is a common oxide of sulfur.

71. Define the problem: Given the percent mass of elements in a compound, determine the empirical formula.

Develop a plan: Choose a convenient sample mass of KxCryOz, such as 100.00 g. Find the masses of K, Cr

and O in the sample, using the given mass percent. Use the molar mass of the elements to find their mo les, then use a whole-number mole ratio to determine the empirical formula.

Execute the plan: 100.0 g of .KxCryOz contains 26.57 g K, 35.36 g Cr, and 38.07 g O.

26 .57 g K ×

1 mol K39.0983 g K

= 0.6796 mol K 35 .36 g Cr ×

1 mol Cr51.9961 g Cr

= 0.6801 mol Cr

38 .07 g O×

1 mol O15.9994 g O

= 2.379 mol O


0.6796 mol K : 0.6801 mol Cr : 2.379 mol O

1 K : 1 Cr : 3.5 O

Multiply by 2 to get whole numbers:

2 K : 2 Cr : 7 O

Use the whole number ratio for the subscripts in the formula. The empirical formula is K2Cr2O7.


Check your answer: The percent by mass of oxygen in the compound is somewhat greater than the percent by mass of potassium or chromium. Since K and Cr atoms weigh more than oxygen atoms, it makes sense that there are several O atom per K and Cr atom present in this formula. This formula coincides with potassium dichromate, a common ionic compound.

72. Define the problem: Given the mass of a compound, the identity of the elements in the compound, and the identity and masses of all the products produced, determine the empirical formula.

Develop a plan: Use the molar mass of the products to find their moles and use the stoichiometry of their formulas to determine the moles of the elements in the compound. Then use a whole-number mo le ratio to determine the empirical formula.

Execute the plan:

The compound is a hydrocarbon and contains only C and H: CiHj. Its combustion produced H2O and CO2. Use molar mass and formula stoichiometry to determine the moles of C and H.

1.481 g CO2 ×

1 mol CO244.0095 g CO2

×1 mol C

1 mol CO2= 0.03365 mol C

0.303 g H2O×

1 mol H2O18.0152 g H2O

×2 mol H

1 mol H2O= 0.0336 mol H


0.03365 mol C : 0.0336 mol H

1 C : 1 H

Use the whole number ratio for the subscripts in the formula. The empirical formula is CH.

Check your answer: These are not necessary calculations, but we can calculate the masses of each element, C and H.

0.03365 mol C ×

12.0107 g C1 mol C

= 0.4041 g C 0.0336 mol H ×1.0079 g H

1 mol H= 0.0339 g H

The sum of these masses (0.4041 g + 0.0339 g = 0.4380 g) adds up to the mass of the original compound (0.438 g). This answer makes sense.


Develop a plan: Use the molar mass of the products to find their moles and use the stoichiometry of their formulas to determine the moles of the elements in the compound. Then use a whole-number mole ratio to determine the empirical formula.

Execute the plan: The compound is a hydrocarbon and contains only C and H: CiHj. Its combustion

produced H2O and CO2. Use molar mass and formula stoichiometry to determine the moles of C and H.

0.379 g CO2 ×

1 mol CO244.0095 g CO2

×1 mol C

1 mol CO2= 8.61 ×10−3 mol C

0.1035 g H2O×

1 mol H2O18.0152 g H2O

×2 mol H

1 mol H2O=1.149 × 10−2 mol H



8.61 × 10–3 mol C : 1.149 × 10–2 mol H

1 C : 1.33 H

Multiply by 3, to get a whole number ratio: 3 C : 4 H

Use the whole number ratio for the subscripts in the formula. The empirical formula is C3H4.

Check your answer: These are not necessary calculations, but we can calculate the masses of each element, C and H.

8.61 ×10−3 mol C×

12.0107 g C1 mol C

= 0.103 g C

0.01149 mol H ×

1.0079 g H1 mol H

= 0.01158 g H

The sum of these masses (0.103 g + 0.01158 g = 0.115 g) add up to the mass of the original compound (0.115 g). This answer make sense.


Develop a plan: Combustion uses oxygen. When the compound also contains oxygen, determine the amount of oxygen after the other elements. Use the molar mass of the products to find their moles and use the stoichiometry of their formulas to determine the moles of the elements that are not oxygen in the compound. Find the masses of those elements, and subtract them from the total mass of the compound to get the mass of oxygen in the compound. Then use the molar mass to calculate the moles of oxygen in the compound. Then use a whole-number mole ratio to determine the empirical formula.

Execute the plan: The compound contains C, H, and O: CiHjOk. Its combustion produced H2O and CO2. Use molar mass and formula stoichiometry to determine the moles of C and H. Use the whole number ratio for the subscripts in the formula.

0.421 g CO2 ×

1 mol CO244.0095 g CO2

×1 mol C

1 mol CO2= 9.56 ×10−3 mol C

0.172 g H2O ×

1 mol H2O18.0152 g H2O

×2 mol H

1 mol H2O=1.91× 10−2 mol H

Calculate the masses of C and H.

9.56 × 10−3 mol C ×

12.0107 g C1 mol C

= 0.115 g C 0.0191 mol H ×1.0079 g H

1 mol H= 0.0192 g H

Calculate the masses of O by subtracting the masses of C and H from the given total compound mass.

0.236 g CiHjOk – 0.115 g C – 0.0192 g H = 0.102 g O

Calculate the moles of O.

0.102 g H2O ×

1 mol O15.9994 g O

= 6.37 × 10−3 mol O



9.56 × 10–3 mol C : 1.91 × 10–2 mol H : 6.37 × 10–3 mol O 1.5 C : 3 H : 1 O

Multiply by 2, to get a whole number ratio: 3 C : 6 H: 2 O

Use the whole number ratio for the subscripts in the formula. The empirical formula is C3H6O2.

Check your answer: The mole ratio came out very close to whole number values. This answer makes sense.


Develop a plan: Combustion uses oxygen. When the compound also contains oxygen, determine the amount of oxygen after the other elements. Use the molar mass of the products to find their moles and use the stoichiometry of their formulas to determine the moles of the elements that are not oxygen in the compound. Find the masses of those elements, and subtract them from the total mass of the compound to get the mass of oxygen in the compound. Then calculate the moles of oxygen in the compound. Then use a whole-number mole ratio to determine the empirical formula.

Execute the plan: The compound contains C, H, and O: CiHjOk. Its combustion produced H2O and CO2. Use molar mass and formula stoichiometry to determine the moles of C and H. Use the whole number ratio for the subscripts in the formula.

0.257 g CO2 ×

1 mol CO244.0095 g CO2

×1 mol C

1 mol CO2= 5.84 ×10−3 mol C

0.0350 g H2O ×

1 mol H2O18.0152 g H2O

×2 mol H

1 mol H2O=3.89 ×10−3 mol H

Calculate the masses of C and H.

5.84 × 10−3 mol C ×

12.0107 g C1 mol C

= 0.0701 g C

3.89 × 10−3 mol H ×

1.0079 g H1 mol H

=0.00392 g H

Calculate the masses of O by subtracting the masses of C and H from the given total compound mass.

0.105 g CiHjOk – 0.0701 g C – 0.00392 g H = 0.031 g O

Calculate the moles of O.

0.031 g H2O×

1 mol O15.9994 g O

= 1.9 ×10−3 mol O


5.84 × 10–3 mol C : 3.89 × 10–3 mol H : 1.9 × 10–3 mol O

3.0 C : 2.0 H : 1.0 O

Use the whole number ratio for the subscripts in the formula. The empirical formula is C3H2O.


Check your answer: The mole ratio came out very close to whole number values. This answer make sense.

General Questions

76. Define the problem: Given the unbalanced chemical equation and the mass of the limiting reactant, determine the mass of a product that can be isolated.

Develop a plan: Balance the equation. Use the molar mass of the reactant to find moles of reactant, then use the stoichiometry of the equation to determine moles of product, then use the molar mass of the product to get the mass of the product.

Execute the plan: Select a balance order: H, O, N, Cu

2 NH3 + 3 CuO N2 + 3 Cu + 3 H2O

Check the balance: 2 N, 6 H, 3 Cu, 3 O

26.3 g NH3 ×

1 mol NH317.0304 g NH3

×1 mol N2

2 mol NH3×

28.0134 g N21 mol N2

=21.6 g N2

Check your answer: It makes sense that the initial mass of NH3 is nearly the same as the mass of N2 created,

since the NH3 molar mass is about half that of N2 and the stoichiometry is 2:1.

77. Define the problem: Given the mass of the only reactant, the formula of the reactant with one element unknown, a balanced equation showing its decomposition, and the mass and identity of one of the products, determine the identity of the unknown element from a given list.

Develop a plan: Use the molar mass of the product to find moles of product, then use the stoichiometry of the equation to determine moles of reactant. Then divide the grams of reactant by the calculated moles of reactant to determine the molar mass of the reactant. Subtract the molar masses of the known elements in the compound to determine the molar mass of the unknown element. Compare the molar masses of the elements on the list and find the one that most closely matches it.

Execute the plan: Use mass of CO2 to find moles of MCO3

0.376 g CO2 ×

1 mol CO244.0095 g CO2

×1 mol MCO31 mol CO2

= 8.54 ×10−3 mol MCO3

Divide given mass by moles:

1.056 g MCO3

8.54 ×10−3 mol MCO3=124

gmol

= molar mass of MCO3

Subtract the molar mass of C and three times the molar mass of O from this molar mass to find the molar mass of M:

124 g MCO3mol MCO3

–

12.0107 g Cmol MCO3

–

3×15.9994 g Omol MCO3

=64 g M

The element in the given list with the closest molar mass to 64 g/mol is (b) Cu.

Check your answer: The molar mass of Cu (63.5 g/mol) is 64. None of the others in the list (Ni at 58.7 g/mol, Zn at 65.4 g/mol, or Ba at 137.2 g/mol) are this close, though within ±1 g/mol Zinc almost qualifies. If we


carry more decimal places than strictly allowed by the rules of significant figures, the molar mass of MCO3 is 123.65 g/mol and the molar mass of M is 63.6 g/mol. This might make us feel better selecting Cu, although these additional significant figures are actually unknown with the limited data provided.

78. Define the problem: Given the balanced chemical equation and the masses of both reactants, determine the maximum yield of a product in grams.

Develop a plan: Find the limiting reactant by determining moles of a product made from each reactant. Then using the molar mass of the product, determine the mass from the moles produced by the limiting reactant.

Execute the plan:

365 g UO3 ×

1 mol UO3286.0271 g UO3

×6 mol UF46 mol UO3

=1.28 mol UF4

365 g BrF3 ×

1 mol BrF3136.8992 g BrF3

×6 mol UF48 mol BrF3

=2.00 mol UF4

UO3 is the limiting reactant, since fewer moles of UF4 are produced from the UO3 reactant (1.28 mol) than

produced from the BrF3 reactant (2.00 mol). Therefore, use 1.28 mol UF4 and the molar mass to determine the

grams of UF4 produced:

1.28 mol UF4 ×

314.0225 g UF41 mol UF4

= 401 g UF4

Check your answer: It makes sense that UO3 is the limiting reactant, because the reactants are present in

equal masses, and UO3 has a larger molar mass than BrF3. Since the molar mass of the product is somewhat larger than the molar mass of the limiting reactant and their stoichiometry is 1:1, it makes sense that slightly larger mass of product is formed in this reaction.

79. Define the problem: Given the balanced chemical equation and the mass of one reactant and the moles of the other reactant, determine the maximum theoretical mass of a product.


Execute the plan:

15.5 g (NH 4)2 PtCl4 ×

1 mol ( NH4)2 PtCl4372.967 g (NH4)2 PtCl 4

×1 mol Pt(NH3)2Cl 21 mol (NH4 )2PtCl 4

= 0.0416 mol Pt(NH3)2Cl2

0.15 mol NH3 ×

1 mol Pt(NH 4)2Cl22 mol NH 3

=0.075 mol Pt( NH4)2Cl 2

(NH4)2PtCl4 is the limiting reactant, since fewer moles of Pt(NH3)2Cl2 are produced from the (NH4)2PtCl4

reactant (0.0416 mol) than produced from the NH3 reactant (0.075 mol). Therefore, use

0.0416 mol Pt(NH3)2Cl2 and the molar mass to determine the grams of Pt(NH3)2Cl2 produced:

0.0416 mol Pt(NH3)2Cl2 ×

300.0448 g Pt(NH 3)2 Cl21 mol Pt(NH3)2Cl2

=12.5 g Pt(NH3)2Cl2


Check your answer: It is satisfying that (NH4)2PtCl4 is the limiting reactant, because that reactant contains the expensive platinum metal and the experimenter would want to make sure all of that got used up. Since the molar mass of the product is somewhat smaller than the molar mass of the limiting reactant and their stoichiometry is 1:1, it makes sense that a slightly smaller mass of product is formed in this reaction.

80. Define the problem: Given the balanced chemical equation, the moles of one reactant, and the mass of the other reactant, determine the maximum yield of a product in grams.


Execute the plan:

2.19 ×10−2 mol H2SO4 ×

1 mol B2H61 mol H2SO4

= 0.0219 mol B2H6

1.55 g NaBH 4 ×

1 mol NaBH 437.8324 g NaBH 4

×1 mol B2H6

2 mol NaBH4= 0.0205 mol B2H6

NaBH4 is the limiting reactant, since fewer moles of B2H6 are produced from the NaBH4 reactant (0.0219 mol)

than produced from the H2SO4 reactant (0.0205 mol). Therefore, use 0.0219 mol B2H6 and the molar mass to

determine the grams of B2H6 produced:

0.0205 mol B2H6 ×

27.6694 g B2H61 mol B2H6

= 0.567 g B2H6

Check your answer: It is satisfying that NaBH4 is the limiting reactant, because that reactant contains the boron element and the experimenter would want to make sure all of that got converted to product. Since the molar mass of the product is somewhat smaller than the molar mass of the limiting reactant and their stoichiometry is 2:1, it makes sense that slightly smaller mass of product is formed in this reaction.



Execute the plan:

The compound is SixOy. Its combustion produced SiO2 and H2O. Use molar mass and formula stoichiometry to determine the moles of Si and H.

11.64 g SiO2 ×

1 mol SiO 260.0843 g SiO2

×1 mol Si

1 mol SiO 2= 0.1937 mol Si

6.980 g H2O ×

1 mol H2O18.0152 g H2O

×2 mol H

1 mol H2O=0.7749 mol H


0.1937 mol Si : 0.7749 mol H

1 Si : 4.000 H


Use the whole number ratio for the subscripts in the formula. The empirical formula is SiH4.

Check your answer: These are not necessary calculations, but we can calculate the masses of each element, Si and H.

0.1937 mol Si ×

28.0855 g Si1 mol Si

= 5.440 g Si 0.7749 mol H ×1.0079 g H

1 mol H= 0.7810 g H

The sum of these masses (5.440 g + 0.7810 g = 6.221 g) add up to the mass of the original compound (6.22 g), to the given significant figures. This answer makes sense.

82. Define the problem: Given the mass of a compound, the identity of the elements in the compound, the mass of one of the products produced, and the identity of all products produced, determine the empirical formula.


Execute the plan:

The compound is BxHy. Its combustion produced B2O3 and H2O. Use molar mass and formula stoichiometry to determine the moles of B.

0.422 g B2O3 ×

1 mol B2O369.6202 g B2O3

×2 mol B

1 mol B2O3= 0.0121 mol B

Find the mass of B in the compound:

0.0121 mol B ×

10.811 g B1 mol B

= 0.131 g B

Find the mass of H in the compound, by subtracting the mass of B from the compound sample mass:

0.148 g BxHy – 0.131 g B = 0.017 g H

Now, determine moles of H:

0.017 g H ×

1 mol H1.0079 g H

= 0.017 mol H


0.0121 mol B : 0.017 mol H

1 B : 1.4 H

If we multiply by 5, we get a whole-number mole ratio: 5 B : 7 H

Use this whole number ratio, the empirical formula is B5H7.

However, the significant figures in the calculation are small, so the quantity of H has an uncertainty of ± 0.1. That means it could also really be a ratio of 1 : 1.5 or 1 : 1.33, and the multipliers would be 2 or 3, respectively to give whole-number mole ratios of:

2 B : 3 H or 3 B : 4 H

They have empirical formulas of B2H3 and B3H4.


Check your answer: The first sentence of the question gives us the impression that many boron hydride compounds exist. The limited significant figures here do not allow us to rule out any of these three empirical formulas: B5H7, B2H3, or B3H4.

83. Define the problem: Given equal moles of all the reactants, determine the limiting reactant. Given equal masses of all the reactants, determine the limiting reaction.

Develop a plan: When the same moles of every reactant is present, the reactant with the largest stoichiometric coefficient is going to be the limiting reactant. In the second part, use the molar mass of each reactant to find their moles, then use the stoichiometry of the reaction to determine the moles of a product formed from each reactant. The reactant that produces the least amount of product is the limiting reactant.

Execute the plan:

First Part: 5 moles of each reactant is present. The stoichiometric coefficients are: 4 for KOH, 2 for MnO2, 1

for O2 and 1 for Cl2. The reactant with the largest stoichiometric coefficient is KOH. So, KOH is the limiting reactant.

Second Part: 5 grams of each reactant is present. Determine the moles of KCl they each form:

5 g KOH ×

1 mol KOH56.1056 g KOH

×2 mol KCl

4 mol KOH= 0.04 mol KCl

5 g MnO2 ×

1 mol MnO286.9368 g MnO2

×2 mol KCl

2 mol MnO2= 0.06 mol KCl

5 g O2 ×

1 mol O231.9988 g O2

×2 mol KCl1 mol O2

=0.3 mol KCl

5 g Cl2 ×

1 mol Cl270.906 g Cl2

×2 mol KCl1 mol Cl2

=0.1 mol KCl

Only 0.04 mol of HCl is produced from KOH, so the KOH is the limiting reactant here, too.

Check your answers: The first part is easy. The reaction says that more moles of KOH are needed that any other reactant, so when equal moles of reactants are present, it runs out first. The differences in molar mass are insufficient to keep the large stoichiometric coefficient for KOH from making it the limiting reactant when equal masses of the reactants are present. These answers make sense.

84. Define the problem: Given equal moles of all the reactants, determine the limiting reactant. Given equal masses of all the reactants, determine the limiting reaction.

Develop a plan: When the same moles of every reactant is present, the reactant with the largest stoichiometric coefficient is going to be the limiting reactant. In the second part, use the molar mass of each reactant to find their moles, then use the stoichiometry of the reaction to determine the moles of a product formed from each reactant. The reactant that produces the least amount of product is the limiting reactant.

Execute the plan:

(a) 10 moles of each reactant is present. The stoichiometric coefficients are: 4 for NaCl, 2 for SO2, 2 for H2O

and 1 for O2. The reactant with the largest stoichiometric coefficient is NaCl. So, NaCl is the limiting reactant.


(b) 100 grams of each reactant is present. (note: 100 has just one significant figure) Determine the moles of KCl they each form:

100 g NaCl ×

1 mol NaCl58.4428 g NaCl

×4 mol HCl

4 mol NaCl= 1.71 mol HCl ≈ 2 mol HCl

100 g SO2 ×

1 mol SO264.0683 g SO2

×4 mol HCl2 mol SO2

= 3.12 mol HCl ≈ 3 mol HCl

100 g H2O ×

1 mol H2O18.0152 g H2O

×4 mol HCl2 mol H2O

=11.1 mol HCl ≈10 mol HCl

100 g O2 ×

1 mol O231.9988 g O2

×4 mol HCl1 mol O2

=12.5 mol HCl ≈10 mol HCl

Only 2 mol of HCl is produced from NaCl, so the NaCl is the limiting reactant here, too.

Check your answers: (a) The reaction says that more moles of NaCl are needed that any other reactant, so when equal moles of reactants are present, it runs out first. (b) The differences in molar mass are insufficient to keep the large stoichiometric coefficient for NaCl from making it the limiting reactant when equal masses of the reactants are present. These answers make sense.

Applying Concepts

85. Two butane molecules react with 13 diatomic oxygen molecules to produce eight carbon dioxide molecules and ten water molecules.

Two moles of gaseous butane molecules react with 13 moles of gaseous diatomic oxygen molecules to produce eight moles of gaseous carbon dioxide molecules and ten moles of liquid water molecules.

86. One P4 molecule reacts with six diatomic chlorine molecules to produce four PCl3 molecules.

One mole of solid P4 molecules reacts with six moles of gaseous diatomic chlorine molecules to produce four

moles of liquid PCl3 molecules.

87. Balance the equation for this reaction: 4 A2 + AB3 3

Reactant number of A atoms = 4 × 2 + 1 = 9 Product number of A atoms = 9 = 3 × 3

Reactant number of B atoms = 3 Product number of A atoms = 3 = 3 × 1

So, the product molecule has 3 A atoms and 1 B atom. Its formula is A3B.

4 A2 + AB3 3 A3B

88. Balance the equation for this reaction: 3 A2B3 + B3 6

Reactant number of A atoms = 3 × 2 = 6 Product number of A atoms = 6 = 6 × 1

Reactant number of B atoms = 3 × 3 + 3 = 12 Product number of A atoms = 12 = 6 × 2

So, the product molecule has 1 A atom and 2 B atoms. Its formula is AB2.

3 A2B3 + B3 6 AB2


89. Define the problem: Given the moles of all the reactants, determine the limiting reactant and the ions present at the end or the reaction.

Develop a plan: Use the stoichiometry of the reaction to determine the moles of a product formed from each reactant. The reactant that produces the least amount of product is the limiting reactant. The excess reactant and the products are present at the end of the reaction. From that information, determine what ions are in the solution after the reaction is complete.

Execute the plan: 1.5 mol Cu ×

1 mol Cu(NO3)21 mol Cu

= 1.5 mol Cu(NO3)2

4.0 mol AgNO3 ×

1 mol Cu(NO3)22 mol AgNO3

= 2.0 mol Cu(NO3)2

Only 1.5 mol Cu(NO3)2 can be formed, so the limiting reactant is Cu and the excess reactant is AgNO3.

That means the solution contains: Ag+ ions, Cu2+ ions, and NO3– ions after the reaction is over. (The

question only asked what ions are present not how many.)

If we want to know how many, we need to determine the moles of AgNO3 actually reacted:

1.5 mol Cu ×

2 mol AgNO31 mol Cu

= 3.0 mol AgNO3 reacted

The moles of AgNO3 left = 4.0 mol AgNO3 initial – 3.0 AgNO3 mol reacted = 1.0 mol AgNO3 left

Quantitatively, the solution has 1.0 mol Ag+ ions, 1.5 mol Cu2+ ions, and (1.0 mol + 2 × 1.5 mol =) 4.0 mol

NO3– ions.

Check your answers: The excess reactant is an ionic compound and one of the products is an ionic compound, so some of the ions present at the beginning are still present when the Cu runs out. The quantities of Ag+ ions have dropped, since some of the silver atoms are now part of the Ag solid product.

The quantity of NO3– ions doesn’t change, since neither N nor O are part of the solid product formed.

These answers make sense.

90. Define the problem: Given the number of molecules of all the reactants, determine what molecules are present after a reaction is complete and determine the limiting reactant.

Develop a plan: Use the stoichiometry of the reaction to determine the molecules of the product formed from each reactant. The reactant that produces the least amount of product is the limiting reactant. The excess reactant and the products are present at the end of the reaction. From that information, determine what figure represents the proper diagram of the post-reaction mixture.

Execute the plan:

6 molecules N2 ×

2 molecules NH31 molecules N2

=12 molecules NH 3

12 molecules H2 ×

2 molecules NH33 molecules H2

= 8 molecules NH3

Only 8 molecules NH3 are formed, because the limiting reactant is H2. The excess reactant is N2.

Determine the number of molecules of N2 that actually reacted:


12 molecules H2 ×

1 molecules N23 molecules H2

= 4 molecules N2 reacted

The molecules of N2 left = 6 molecules of N2 initial – 4 molecules of N2 reacted = 2 molecules of N2 left

These results indicate that Figure 4 is the right figure, and that (b) is the right statement regarding the limiting reactant: “H2 is the limiting reactant”

Check your answers: There are two molecules of N2 left after all the molecules of H2 react, so it makes

sense that H2 is the limiting reactant. The figure shows 8 molecules of NH3 and 2 molecules of N2. These answers make sense.

91. The reactants are six O2 molecules and six CO molecules. The given equation described s 1:2 reaction, so we

determine that the CO molecules run out first, after producing six CO2 molecules and using three O2

molecules. That leaves three O2 molecule unreacted:

Note: The black circles are carbon, and the white circles are oxygen.

92. Four XY3 molecules are made from two diatomic X2 molecules and six diatomic Y2 molecules. So,

symbolically, the reaction is 2 X2 + 6 Y2 4 XY3, and the stoichiometric equation representing that

reaction is (b) X2 + 3 Y2 2 XY3

93. The increase of NaHCO3 causes an obvious increase in the amount of gas produced, as indicated by the larger volume of the balloon in #1 through #4. In #5 and #6, the volume look unchanged, but there are subtle differences in the shape and “droopiness” of the balloon, indicating a possible increase in the pressure inside the balloon. If we use volume alone, then the NaHCO3 limits up to #4, then the acetic acid is the limiting reactant in #5 and #6. If we consider an increase in pressure as an indication of more gas being produced in each balloon, then we would conclude that the NaHCO3 was limiting in every experiment.

94. When masses smaller than 1.2 gram of the metal are added, the metal is the limiting reactant, so the mass of the compound produced is directly proportional to the mass of metal present (shown by the straight line rising up to the right on the graph). More metal makes more products when the mass of metal is less than 1.2 gram.

When the mass larger than 1.2 g of metal are added, the bromine is the limiting reactant, so the particular mass of the metal is independent of how much compound is made. Since the amount of bromine is held constant, the mass of compound formed is also a constant (shown by a horizontal line on the graph).

Conclusion: When metal mass is less than 1.2 g the metal is the limiting reactant. When the metal mass is greater than 1.2 g the bromine is the limiting reactant.


95. Define the problem: Given the graph of a product’s mass vs. a metal element reactant’s mass, the identity of the elemental reactants and the information that one compound is formed, determine the empirical formula of the product compound, write the balanced equation for the reaction, and name the product.

Develop a plan: The point where the graph levels out is the stoichiometric equivalence point. (See the answer to Question 92 in the Student Solution Manual for details). Assuming 100 % yield, subtract the metal mass from the product mass to get the other reactant’s mass. Use the molar mass of the reactant elements to find their moles. Then use a whole-number mole ratio to determine the empirical formula. Write and balance the equation and name the product.

Execute the plan: Estimating the location of the stoichiometric equivalence point, we find that 12.6 grams of compound are made from 2.3 grams of Fe.

12.6 g compound – 2.3 g Fe = 10.3 g Br

The compound is a contains only Fe and Br: FeiBrj.

2.3 g Fe ×

1 mol Fe55.845 mol Fe

= 0.041 mol Fe 10 .3 g Br ×

1 mol Br79.904 mol Br

= 0.13 mol Br


0.041 mol Fe : 0.13 mol Br

1 Fe : 3 Br

Use the whole number ratio for the subscripts in the formula. The empirical formula is FeBr3. Remembering that bromine is a diatomic liquid element and that iron is a monatomic solid, the equation looks like this: 2 Fe(s) + 3 Br2(l) 2 FeBr3(s)

The compound is iron(III) bromide.

Check your answers: The approximations that had to be made from the graph provide a wide rage of values, depending on a persons estimations.. However, the result is reasonable, due to the fact that iron has two typical oxidation states, +2 and +3, and that bromine has only one, –1. There is no way the data could support the answer being iron(II) bromide.

More Challenging Problems

96. Define the problem: Given the formulas of the reactants and one product and the percent mass of elements in the second product, determine the balanced chemical equation.

Develop a plan: Choose a convenient sample mass of product, FexOy, such as 100.0 g. Find the mass of Fe

and O in the sample, using the given mass percent. Use the molar mass of the elements to find their moles, then use a whole-number mole ratio to determine the empirical formula. Using the formulas of the reactants and products balance the equation.

Execute the plan: 100.0 g of .FexOy contains 72.3 g Fe and 27.7 g O.

72 .3 g Fe ×

1 mol Fe55.845 g Fe

= 1.29 mol Fe 27 .7 g O×

1 mol O15.9994 g O

= 1.73 mol O


1.29 mol Fe : 1.73 mol O


1 Fe : 1.34 O


3 Fe : 4 O

Use the whole number ratio for the subscripts in the formula. The empirical formula is Fe3O4.

? H2(g) + ? Fe2O3(s) ? H2O(l) + ? Fe3O4(s) Select order: Fe, O, H

? H2(g) + 3 Fe2O3(s) ? H2O(l) + 2 Fe3O4(s) 6 Fe

? H2(g) + 3 Fe2O3(s) 1 H2O(l) + 2 Fe3O4(s) 9 O

1 H2(g) + 3 Fe2O3(s) 1 H2O(l) + 2 Fe3O4(s) 2 H

Check your answer: Fe3O4 is a common oxide of iron. 6 Fe, 9 O and 2 H on each side.

97. Define the problem: Given the percent mass of elements in an organic compound, determine the balanced chemical equation for the combustion reaction.

Develop a plan: Choose a convenient sample mass of product, CxHyOz, such as 100.00 g. Find the mass of

C and H in the sample, using the given mass percent, then subtract those masses from the total sample mass to get the mass of O. Use the molar mass of the elements to find their moles, then use a whole-number mole ratio to determine the empirical formula. For combustion, use the formula of the organic compound and O2

as reactants and CO2 and H2O as products, then balance the equation.

Execute the plan: 100.00 g of .CxHyOz contains 34.62 g C and 3.88 g H.

100.00 g of .CxHyOz – 34.62 g C – 3.88 g H = 61.50 g O

34 .62 g C×

1 mol C12.0107 g C

= 2.882 mol C 3.88 g H ×

1 mol H1.0079 g H

= 3.85 mol H

61 .50 g O ×

1 mol O15.9994 g O

= 3.844 mol O


2.882 mol C : 3.85 mol H : 3.844 mol O

1 C : 1.336 H : 1.334 O


3 C : 4 H : 4 O


? C3H4O4 + ? O2 ? CO2 + ? H2O Select order: C, H, O

1 C3H4O4 + ? O2 3 CO2 + ? H2O 3 C

1 C3H4O4 + ? O2 3 CO2 + 2 H2O 4 H


1 C3H4O4 + 2 O2 3 CO2 + 2 H2O 8 O

Check your answer: The mole ratio is quite close to whole number values. 3 C, 4 H, and 8 O on each side.

98. Define the problem: Given a balanced chemical equation, the volume of a liquid limiting reactant, and the density of the liquid, determine the maximum theoretical yield of a product.

Develop a plan: Use the density of the reactant to find the grams of the reactant. Then use the molar mass of the reactant to find the moles of reactant. Then use the stoichiometry of the equation to determine moles of product. Then use the molar mass of the product to get the mass of the product.

Execute the plan:

25.0 mL Br2 ×

3.1023 g Br21 mL Br2

×1 mol Br2

159.808 g Br2×

1 mol Al2Br63 mol Br2

×533.387 g Al2Br6

1 mol Al2Br6= 86.3 g Al2Br6

Check your answer: The units all cancel, and the numerical multipliers are larger than the dividers. This somewhat larger answer makes sense.

99. Define the problem: Given the formula of two binary oxide compounds both containing an unknown element and an unknown amount of oxygen, and given the mass of the both elements in a given sample of each compound, we need to determine the ratio of oxygen atoms in the two formulas. Given the number of oxygen atoms in one formula, determine the identity of the unknown element.

Develop a plan: We have incomplete information in this problem, so we’ll let the ratio help us eliminate the need to know information common between the two substances. Start by converting the mass of O in each sample to moles of O. We then use the mass of A and its molar mass to determine moles of A, then use the formula stoichiometry to determine the moles of AOx. Dividing the moles of O by the moles of AOx tells us the value of x. Follow the same procedure to get y. We don’t know molar mass of A, but it will cancel in the ratio.

Execute the plan:

(a) Find moles of O: 3.2 g O ×

1 mol O15.9994 g O

= 0.20 mol O

Find moles of AOx, using MA for the molar mass of unknown element A:

1.2 g A ×

1 mol AMA g A

×1 mol AOx

1 mol A=

1.2M A

mol AOx

Find moles of AOy in a similar fashion:

2.4 g A ×

1 mol AM A g A

×1 mol AOy

1 mol A=

2.4M A

mol AOy

Now set up the ratio to get

xy

:

mol Omol AOx

mol Omol AOy


0.20 mol O1.2MA

mol AOx

0.20 mol O2.4MA

mol AOy

=

2 mol O1 mol AOx

1 mol O1 mol AOy

xy

=21

(b) If x = 2, we can use the moles of O to tell use how many moles of A are present in the AOx sample.

0.20 mol O×

1 mol A2 mol O

= 0.10 mol A

Divide mass of A in the sample by moles of A in the sample to get molar mass of A:

1.2 g A0.10 mol A

= 12 g / mol The element with this molar mass is carbon, C.

Check your answers: (a) The law of multiple proportions assures us that the ratio of elements in binary compounds is a small whole number, so the ratio looks right. There are probably many other legitimate ways to solve this problem. (b) The molar mass of the unknown element was very close to the molar mass of carbon, and both CO and CO2 are known compounds, so this result makes sense also.


Develop a plan: (a) Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. (b) Use the moles of product produced from the limiting reactant and the molar mass of the product to find the grams. (c) From the quantity of limiting reactant, determine the moles of the other reactant needed for complete reaction. Convert that number to grams using the molar mass. Then subtract the quantity used from the initial mass given to get the mass of excess reactant.

Execute the plan: Note that 500 grams has 1 significant figure and 1300 grams has 2 significant figures.

(a) The balanced equation says: 1 mol CH4 produces 3 mol H2.

500 g CH4 ×

1 mol CH416.0423 g CH 4

×3 mol H2

1 mol CH 4= 90 mol H2

The balanced equation says: 1 mol H2O produces 3 mol H2.

1300 g H2O ×

1 mol H2O18.0152 g H2O

×3 mol H21 mol H2O

= 220 mol H2

The number of H2 moles produced from CH4 is smaller (90 mol < 220 mol), so CH4 is the limiting reactant and H2O is the excess reactant.

(b) Find the mass H2: 90 mol H2 ×

2.0158 g H21 mol H2

=200 g H2


(c) 500 g CH4 ×

1 mol CH416.0423 g CH 4

×1 mol H2O1 mol CH 4

×18.0152 g H2O

1 mol H2O= 600 g H2O

1300 g H2O initial – 600 g H2O used up = 700 g H2O remains unreacted

Check your answers: The small number of significant figures makes the uncertainty in this calculation somewhat high. However, the general expectations are still met. There is a larger mass of H2O present, and the molar masses of the reactants are about the same size. Since the equation indicates that equal moles of H2O and CH4 react, it makes sense that CH4 is the limiting reactant. In addition, the calculation in (c) proves

that the initial mass of H2O was larger than required to react with all of the CH4.

101. Define the problem: Given the mass of a sample of ore with unknown amounts of two copper compounds and some amo unt of inert impurity, the products of a chemical reaction, and the mass and percent purity of one element in the products, determine the percent by mass of one of the copper compounds.

Develop a plan: Use the percent purity of the copper in the product and the molar mass of copper to find the moles of copper in the ore. Set two variables X = the mass in grams of Cu2S and Y = the mass in

grams of CuS. Then establish two equations, one describing the total mass of the sample related to the three components and one describing the moles of copper in the sample from two sources. Algebraically solve for X and Y, then use the mass of Cu2S and the total sample mass to find the percentage by mass.

Execute the plan: 150.8 g impure Cu ×

90.0 g Cu100 gimpure Cu

×1 mol Cu

63.546 g Cu= 2.136 mol Cu

Let X = grams of Cu2S and Y = grams of CuS, and recall that 10.% of the ore is inert impurities.

Total mass of sample = 200.0 g = X + Y + 0.10 × 200.0 g

Moles of copper =

2.136 mol Cu = X g Cu2S ×

1 mol Cu2S159.157 g Cu2S

×2 mol Cu

1 mol Cu2S+ Y g CuS×

1 mol CuS95.611 g CuS

×1 mol Cu

1 mol CuS

180.0 = X + Y and 2.136 = 0.0125662 X + 0.010459 Y

Solve for X: 2.136 = 0.0125662 X + 0.010459 (180.0 − X)

2.136 = 0.0125662 X + 1.883 − 0.010459 X

2.136 − 1.883 = (0.0125662 − 0.010459) X

0.253 = (0.002107) X

120. = X

120. g Cu2O200.0 g ore

× 100 % = 60.1 % Cu2O

Check your answer: Close to the maximum quantity of iron was produced, so it makes sense that the percent yield is over ninety percent.

102. Define the problem: Given the names of products and reactants of a chemical reaction, and the masses of the two reactants, determine the masses of all the products and reactants after the reaction is complete.


Develop a plan: Determine the formulas of the reactants and products and set up and balance the chemical equation. Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. Use the moles of product produced from the limiting reactant, the molar masses of the products to find the grams of products. Use the moles of product produced, determine the moles of the other reactant needed for complete reaction. Convert that number to grams using the molar mass. Then subtract the quantity used from the initial mass given to get the mass of excess reactant.

Execute the plan: Silver nitrate is AgNO3. Sodium carbonate is Na2CO3. Silver carbonate is Ag2CO3.

Sodium nitrate is NaNO3.

? AgNO3 + ? Na2CO3 ? Ag2CO3 + ? NaNO3 Order: Ag, Na, N, C, O

2 AgNO3 + ? Na2CO3 1 Ag2CO3 + ? NaNO3 2 Ag

2 AgNO3 + 1 Na2CO3 1 Ag2CO3 + 2 NaNO3 2 Na, 2 N, 1 C, 9 O

The balanced equation says: 1 mol Na2CO3 produces 1 mol Ag2CO3.

12.43 g Na 2CO3 ×

1 mol Na 2CO3105.9885 g Na 2CO3

×1 mol Ag2CO31 mol Na 2CO3

= 0.1173 mol Ag2CO 3

The balanced equation says: 2 mol AgNO3 produces 1 mol Ag2CO3.

8.37 g AgNO3 ×

1 mol AgNO3169.8731 g AgNO3

×1 mol Ag2CO 32 mol AgNO3

= 0.0246 mol Ag2CO 3

The numb er of Ag2CO3 moles produced from AgNO3 is smaller (0.0246 mol < 0.1173 mol), so AgNO3 is

the limiting reactant and Na2CO3 is the excess reactant. Therefore, at the end of the reaction, the mass of

AgNO3 present is zero grams.

Find the mass Ag2CO3: 0.0246 mol Ag2CO 3 ×

275.7453 g Ag2CO 31 mol Ag2CO3

=6.79 g Ag2CO3

Find the mass NaNO3: 0.0246 mol Ag2CO 3 ×

2 mol NaNO 31 mol Ag2CO3

×84.9947 g NaNO 3

1 mol NaNO 3= 4.19 g NaNO3

Find the mass Na2CO3 used up, then subtract from intial for mass unreacted:

0.0246 mol Ag2CO 3 ×

1 mol Na 2CO31 mol Ag2CO3

×105.9885 g Na 2CO 3

1 mol Na 2CO3=2.61 g Na 2CO 3 usedup

12.43 g Na2CO3 initial – 2.61 g Na2CO3 used up = 9.82 g Na2CO3 remains unreacted

Check your answers: The total mass before the reaction = 12.43 g + 8.37 g = 20.80 g is equal to the total mass after the reaction = 6.79 g + 4.19 g + 9.82 g = 20.80 g, in accordance with the conservation of mass.

103. Define the problem: Given a balanced equation for a chemical reaction, and the masses of the three reactants, determine the limiting reactant, the mass of one of the product and the masses of the excess reactants after the reaction is complete.


Develop a plan: Use the molar mass of the reactants to find the moles of the reactant substances. Then use the stoichiometry of the equation to determine the moles of the product produced in each case. Identify the limiting reactant from the reactant that produces the least number of products. Use the moles of product produced and the molar mass of the product to find the grams of products. Use the moles of product produced, determine the moles of the other reactants needed for complete reaction. Convert those numbers to grams using the molar masses. Then subtract the quantity used from the initial mass given to get the mass of excess reactant.

Execute the plan:

The balanced equation says: 2 mol SO2 produces 2 mol H2SO4.

200. g SO2 ×

1 mol SO264.064 g SO2

×2 mol H2SO4

2 mol SO2= 3.12 mol H2SO4

The balanced equation says: 1 mol O2 produces 2 mol H2SO4.

85 g O2 ×

1 mol O231.9988 g O2

×2 mol H2SO4

1 mol O2= 5.3 mol H2SO4

The balanced equation says: 1 mol H2O produces 2 mol H2SO4.

66 g H2O×

1 mol H2O18.0152 g H2O

×2 mol H2SO4

2 mol H2O= 3.7 mol H2SO4

The number of H2SO4 moles produced from SO2 is smaller (3.12 mol < 3.7 mol < 5.3 mol), so SO2 is the

limiting reactant.

Find the mass H2SO4: 3.12 mol H2SO4 ×

98.078 g H2SO41 mol H2SO4

= 306 g H2SO4

Find the mass H2O used up, then subtract from intial for mass unreacted:

200. g SO2 ×

1 mol SO264.064 g SO2

×2 mol H2O2 mol SO2

×18.0152 g H2O

1 mol H2O= 56.2 mol H2O used up

66 g H2O initial – 56.2 g H2O used up = 10. g H2O remains unreacted

Find the mass H2O used up, then subtract from intial for mass unreacted:

200. g SO2 ×

1 mol SO264.064 g SO2

×1 mol O22 mol SO2

×31.9988 g O2

1 mol O2= 49.9 mol O2 used up

85 g O2 initial – 49.9 g O2 used up = 35 g O2 remains unreacted

Check your answers: The total mass before the reaction = 200. g + 85 g + 66 g = 351 g is equal to the total mass after the reaction = 306 g + 10. g + 35 g = 351 g, in accordance with the conservation of mass.

104. Define the problem: Given the mass of a sample of a liquid containing unknown amounts of two organic compounds and given the mass of one of the products of a chemical reaction, determine the composition of the liquid.


Develop a plan: First, balance the combustion reactions for ethyl alcohol and methyl alcohol. Set two variables X = the mass in grams of methyl alcohol and Y = the mass in grams of ethyl alcohol. Then establish two equations, one describing the total mass of the sample related to the two components and one describing the moles of carbon dioxide produced when burning the sample. Algebraically solve for X and Y, then use those masses and the total sample mass to find the percentage by mass of the two compounds.

Execute the plan:

Unbalanced: ? C2H5OH(l) + ? O2(g) ? CO2(g) + ? H2O(l) Order: C, H, then O

1 C2H5OH(l) + ? O2(g) 2 CO2(g) + ? H2O(l) 2 C’s

1 C2H5OH(l) + ? O2(g) 2 CO2(g) + 3 H2O(l) 6 H’s

1 C2H5OH(l) + 6 O2(g) 2 CO2(g) + 3 H2O(l) 7 O’s

Unbalanced: ? CH3OH(l) + ? O2(g) ? CO2(g) + ? H2O(l) Order: C, H, then O

1 CH3OH(l) + ? O2(g) 1 CO2(g) + ? H2O(l) 1 C’s

1 CH3OH(l) + ? O2(g) 1 CO2(g) + 2 H2O(l) 4 H’s

1 CH3OH(l) + 32

O2(g) 1 CO2(g) + 2 H2O(l) 4 O’s

Let X = grams of ethyl alcohol and Y = grams of methyl alcohol

Total mass of sample = 0.280 g = X + Y

Moles of carbon dioxide produced when burned = 0.385 g CO2 ×

1 mol CO244.0095 g CO2

= 0.00875 mol CO2

= X g C2H5OH ×

1 mol C2H5OH46.0682 g C2H5OH

×2 mol CO2

1 mol C2H5OH

+ Y g CH 3OH ×

1 mol CH 3OH32.0417 g CH 3OH

×1 mol CO 2

1 mol CH3OH

0.280 = X + Y and 0.00875 = 0.0434139 X + 0.0312093 Y

Solve for X: 0.00875 = 0.0434139 X + 0.0312093(0.280 – X)

0.00875 = 0.0434139 X + 0.000874 − 0.0312093 X

0.00875 − 0.000874 = (0.0434139 − 0.0312093) X

0.00001 = 0.0122046 X

X = 0.0008 g ethyl alcohol

Y = 0.280 – X = 0.280 – 0.0008 = 0.279 g methyl alcohol

0.279 g CH 3OH0.280 g liquid

×100 % = 99.7 % CH 3OH

0.001 g C2H5OH0.280 g liquid

× 100 % = 0.3 % C2H5OH


Check your answer: Assuming the liquid is pure methyl alcohol gives the following mass of CO2:

0.280 g CH 3OH ×

1 mol CH 3OH32.0417 g CH 3OH

×1 mol CO 2

1 mol CH3OH×

44.0095 g CO 21 mol CO2

= 0.385 g CO2

So it makes sense that the liquid is almost one hundred percent methanol.

105. Define the problem: Given the percent mass of elements in an organic compound, determine the balanced chemical equation for the combustion reaction.

Develop a plan: Choose a convenient sample mass of product, CxHyOz, such as 100.00 g. Find the mass

of C and H in the sample, using the given mass percent, then subtract those masses from the total sample mass to get the mass of O. Use the molar mass of the elements to find their moles, then use a whole-number mole ratio to determine the empirical formula. For combustion, use the formula of the organic compound and O2 as reactants and CO2 and H2O as products, then balance the equation.

Execute the plan: 100.00 g of .CxHyOz contains 49.31 g C and 6.90 g H.

100.00 g of .CxHyOz – 49.31 g C – 6.90 g H = 43.79 g O

49.31 g C×

1 mol C12.0107 g C

= 4.106 mol C 6.90 g H ×

1 mol H1.0079 g H

= 6.85 mol H

43.79 g O×

1 mol O15.9994 g O

= 2.737 mol O


4.106 mol C : 6.85 mol H : 2.737 mol O

1.5 C : 2.5 H : 1 O


3 C : 5 H : 2 O


? C3H5O2 + ? O2 ? CO2 + ? H2O Select order: H, C, O

2 C3H5O2 + ? O2 ? CO2 + 5 H2O 10 H

2 C3H5O2 + ? O2 6 CO2 + 5 H2O 6 C

2 C3H5O2 +

132

O2 6 CO2 + 5 H2O 17 O

4 C3H5O2 + 13 O2 12 CO2 + 10 H2O

Check your answer: The mole ratio is quite close to whole number values. 12 C, 20 H, and 34 O on each side.

106. Define the problem: Given the percent mass of elements in an organic compound and the compounds molar mass, determine the empirical formula and the molecular formula.


Develop a plan: Choose a convenient sample mass of product, CxHyNzOw, such as 100.00 g. Find the

mass of C and H in the sample, using the given mass percent, then subtract those masses from the total sample mass to get the mass of O. Use the molar mass of the elements to find their moles, then use a whole-number mole ratio to determine the empirical formula. For combustion, use the formula of the organic compound and O2 as reactants and CO2 and H2O as products, then balance the equation.

Execute the plan:

(a) 100.00 g of .CxHyNzOw contains 54.82 g C, 7.10 g N, and 32.46 g O.

100.00 g of .CxHyNzOw – 54.82 g C – 7.10 g N – 32.46 g O = 5.62 g H

54 .82 g C×

1 mol C12.0107 g C

= 4.564 mol C 7.10 g N ×

1 mol N14.0067 g N

= 0.507 mol N

32 .46 g O ×

1 mol O15.9994 g O

= 2.029 mol O 5.62 g H ×

1 mol H1.0079 g H

= 5.58 mol H


4.564 mol C : 5.58 mol H : 0.507 mol N : 2.029 mol O

9 C : 11 H : 1 N : 4 O

Use the whole number ratio for the subscripts in the formula. The empirical formula is C9H11NO4.

(b) The molar mass of the empirical formula = 197.1875 g/mol C9H11NO4

197.19 g / mol compound197.1875 g / mol emp . formula

= 1 emp . formula / compound

The molecular formula is C9H11NO4.

Check your answers: The mole ratio is quite close to whole number values. The empirical formula is very close to the molecular formula. These answers make sense.

Documents

Chapter 4: Quantities of Reactants and Productsacademic.udayton.edu/VladimirBenin/123_2004/Problem... · 2013. 12. 16. · reactants must add up to the masses of the products, 92.5665