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Stoichiometry
Stoichiometry
• Needs a balanced equation
• Use the balanced equation to predict ending and / or starting amounts
• Coefficients are now mole ratios
In terms of Moles• The coefficients tell us how many moles of
each kind.
• Mole ratio - conversion ratio that relates the amounts in moles of any two substances in a chemical reaction.
• Molar mass - mass, in grams, of one mole of a substance.
ReviewReview: Molar Mass: Molar MassA substance’s A substance’s molar mass molar mass (molecular weight) is the mass in (molecular weight) is the mass in grams of one mole of the compound.grams of one mole of the compound.
COCO22 = 44.01 grams per mole = 44.01 grams per mole
HH22O = 18.02 grams per moleO = 18.02 grams per mole
Ca(OH)Ca(OH)22 = 74.10 grams per mole = 74.10 grams per mole
Mole Ratios
2 Al2O3(l) 4 Al(s) + 3 O2(g)
Mole Ratios Mole Ratio (Fraction)
2 mol Al2O3 : 4 mol Al
2 mol Al2O3 : 3 mol O2
4 mol Al : 3 mol O2
2 mol Al2O3
4 mol Al
2 mol Al2O3
3 mol O2
4 mol Al
3 mol O2
III. Stoichiometric “road map” (Use the balanced chemical equation)
aA + bB cC + dD
Mass A
AtomsMoleculesFormula Units
A
Mol A Mol B
Mass B
AtomsMoleculesFormula Units
B
Mol Ratio Using the Coefficients from
the balanced chemical equation
Molar mass
6.022 x 1023
Molar
mass
6.022 x 10 23
3 A + 4 B 2 D + 1 F
How many moles of F are produced from 1.00 mol of A?
1 mol A
3 mol A
1 mol F
Molecules
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
= 0.33 mol F
How many moles of D are produced from 5.00 mol of B?
5 mol B
4 mol B
2 mol D
= 2.50 mol D
How many moles of lithium carbonate are produced when 5.3 mol CO2 are reacted?
CO2(g) + 2LiOH(s) Li2CO3(s) + H2O(l)
1. What is your starting point?2. What is your ending point?
5.3 mol of CO2
mol of Li2CO3
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
5.3 mol CO2
1 mol CO2
1 mol Li2CO3
= 5.3 mol Li2CO3
3 A + 4 B 2 D + 1 FHow many grams of F are produced from 1.00 mol of A? If MM of F is 10.0g/mol.
1 mol A
3 mol A
1 mol F= 3.33g F
How many grams of D are produced from 5.00 mol of B? MM of D is 20.0g/mol
5 mol B
4 mol B
2 mol D
= 50.0g D
1 mol F
10 g F
1 mol D
20 g D 6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
What is the mass of glucose (C6H12O6) produced from 3.00 mol of water (H2O)?
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
6.022 x 10 23
Mass A
Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar mass
Mola
r mass
6.022 x
1023
Mol Ratio
Molecules
3 mol H2O
1. What is your starting point?2. What is your ending point?
3.00 mol of H2O
g of C6H12O6
6 mol H2O
1 mol C6H12O6
1 mol C6H12O6
=90.1 g C6H12O6
180.81g C6H12O6
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
What is the mass of oxygen (O2) produced from 2.50 mol of water (H2O)?
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
1. What is your starting point?2. What is your ending point?
2.50 mol of H2O
g of O2
2.5 mol H2O
6 mol H2O
6 mol O2
1 mol O2
32.0 g O2
=80.0 g O2
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
How many moles of NO are formed from 824 g of NH3?
1. What is your starting point?2. What is your ending point?
824 g of NH3
mol of NO
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
824 g NH3
17.03g
1 mol NH3
4 mol NH3
4 mol NO
= 48.4 mol NO
3 A + 4 B 2 D + 1 FHow many grams of F are produced from 5.00g of A? If MM of F is 10.0g/mol and MM of A is 25.0g/mol
1 mole A
3 mole A
1 mole F= 0.67g F
How many grams of D are produced from 5.00g of B? MM of D is 20.0g and MM of B is 10.0g/mol
1 mole B
4 mole B
2 mole D
=5.00g D
1 mole F
10 g F
1 mole D
20 g D
6.022 x 10 23
Mass A
Mass B
AtomsMolecules
AAtoms
B
MolA MolB
Molar mass
Mola
r mass
6.022 x
1023
Mol Ratio
Molecules
5 g A
25 g A
5 g B
10 g B
Sn(s) + 2HF(g) SnF2(s) + H2(g)
How many grams of SnF2 are produced from the reaction of 30.00 g HF?
1. What is your starting point?2. What is your ending point?
30.00 g of HF
g SnF2
1 mol HF
2 mol HF
1 mol SnF2
= 117.5g SnF2
1 mol SnF2
156.71 g SnF2 30.00g HF
20.01g HF
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
Working a Stoichiometry ProblemWorking a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
1. Identify reactants and products and write the balanced equation.
Al + O2 Al2O3
b. What are the reactants?
a. Every reaction needs a yield sign!
c. What are the products?
d. What are the balanced coefficients?
4 3 2
=6.50 g Al
? g Al2O3
1 mol Al
26.98 g Al 4 mol Al
2 mol Al2O3
1 mol Al2O3
101.96 g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4
= 12.3 g Al2O3
1. What is your starting point? 6.50 g of Aluminum
2. What is your ending point? g of aluminum oxide
Al + O2 Al2O34 3 2
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
Sn(s) + 2HF(g) SnF2(s) + H2(g)
How many grams of HF are produced from the reaction of 150.5 g H2?
6.022 x 10 23
Mass A Mass B
Atoms
MoleculesA AtomsB
MolA MolB
Molar massMolar
mass
6.022 x 1023
Mol Ratio
Molecules
1 mol H2
1 mol H2
2 mol HF
1 mol HF
20.01g HF 150.5 g H2
2.02g H2
= 2982 g HF
II. Limiting ReagentA. Stoichiometric amounts: The proportions indicated in
the _________ rxn.B. Most reactions do not have stoichiometric amounts.
Generally, one reactant will be _________ before the other. The reactant that is depleted first is known as the ___________________. The reactant that is left at the end of the reaction is called the ___________________.
C. Analogy: How to make a cheese sandwich. 2 slices of bread + 1 slice of cheese → 1 cheese sandwich
If you have 8 slices of bread and 6 slices of cheese, how many sandwiches can you make? __ (theoretical yield)
What is the limiting reagent? ______ What is the excess reagent? _______ How much of the excess reagent is left at the end of the
rxn? ______________
balanced
depleted
limiting reagent (LR)excess reagent (ER)
4bread
cheese
2 slices cheese
D. Theoretical yield: The amount of product in grams that forms if all of the ________________ has reacted. (This number is CALCULATED on paper! Units are in grams only!)
E. Actual yield: The amount of product (in
grams) that is actually made (This number is from the EXPERIMENT).
F. Percent yield: The comparison of the actual yield to the theoretical yield.
limiting reagent
100 x yield lTheoretica
yield Actual yieldPercent
Zn + 2 HCl ZnCl2 + H2
If you have 1 mol of Zn, how much H2 would you make?
If you have 1 mol of HCl, how much H2 would you make?
1 mol Zn
1 mol Zn
1 mol H2= 1 mol H2
1 mol HCl
2 mol HCl
1 mol H2= 0.5 mol H2
What is the limiting reagent?
HCl
How much H2 is produced?
0.5 mol – theoretical yield
Zn + 2 HCl ZnCl2 + H2
If you have 0.25 mol of Zn, how much H2 would you make?
If you have 1.00 mol of HCl, how much H2 would you make?
0.25 mol Zn
1 mol Zn
1 mol H2= 0.25 mol H2
1.00 mol HCl
2 mol HCl
1 mol H2= 0.5 mol H2
What is the limiting reagent? Zn
How much H2 is produced?
0.25 mol - theoretical yield
PCl3 + 3 H2O H3PO3 + 3 HCl3.00 mol PCl3 and 3.00 mol H2O react. Determine the limiting reactant and theoretical yield of HCl.
1. Determine the limiting reactant
3.00 mol H2O
3 mol H2O
3 mol HCl= 3.00 mol HCl
3.00 mol PCl3
1 mol PCl3
3 mol HCl= 9.00 mol HCl EXCESS
LIMITING
2. Determine the theoretical yield of HCl 3.00 mol
PCl3 + 3 H2O H3PO3 + 3 HCl
2. Determine the theoretical yield of HCl 3.00 mol
3. Determine the theoretical yield of HCl in grams
3.00 mol HCl
1 mol HCl
36.46g HCl= 109 g HCl
PCl3 + 3 H2O H3PO3 + 3 HClDetermine the limiting reactant and theoretical yield (g) of H3PO3 if 225 g of PCl3 and 123 g of H2O are reacted.
1. Determine the limiting reactant
1 mol PCl3
1 mol PCl3
1 mol H3PO3
1 mol H3PO3
82.00g H3PO3 225 g PCl3
137.32g PCl3
= 134g H3PO3
1 mol H2O
3 mol H2O
1 mol H3PO3
1 mol H3PO3
82.00g H3PO3 123 g H2O
18.02g H2O
= 187g H3PO3
EXCESS
LIMITING
100 x yield lTheoretica
yield Actual yieldPercent
PCl3 + 3 H2O H3PO3 + 3 HClThe theoretical yield of this reaction is 134g H3PO3. However, the actual yield from the experiment is 120g. Calculate the percent yield.
120 g H3PO4
134 g H3PO4
X 100% = 89.6 %
N2 + 3 H2 2 NH3
Determine the limiting reagent, the theoretical yield and the percentage yield if 14.0g N2 are mixed with 9.0g H2 and the 16.1g NH3 actually formed.
1 mol N2 2 mol NH3
1 mol NH3
17.04g NH3 14 g N2= 17.0g NH3
LIMITING
28.01g N2 1 mol N2
1 mol H2 2 mol NH3
1 mol NH3
17.04g NH3 9 g H2= 50.6g NH3
2.02g H2 3 mol H2
EXCESS
16.1 g NH3X 100% = 94.2 %
17.0 g NH3
16.1g of bromine are mixed with 8.42g of chlorine to give an actual yield of 21.1g of bromine monochloride. Determine the limiting reactant and the percentage yield.
1 mol Br2 2 mol BrCl
1 mol BrCl
115.35g BrCl16.1 g Br2= 23.2g BrCl
LIMITING
159.8g Br2 1 mol Br2
EXCESS
21.1 g BrClX 100% = 90.9 %
23.2 g BrCl
Br2 + Cl2 2 BrCl
1 mol Cl2 2 mol BrCl
1 mol BrCl
115.35g BrCl8.42 g Cl2= 27.4g BrCl
70.9g Cl2 1 mol Cl2
