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StoichiometryStoichiometry – Study of quantitative relationships that can be derived
from chemical formulas and chemical equationsMole ─ Mole Relationship
need a balanced equationMole Ratio – the ratio of moles of one substance to moles of another
substance in a balanced chemical equationThe coefficients in a balanced equation give the relative numbers of
molecules, as well as, the relative number of moles.
CO(g) + 2H2(g) CH3OH(l)
Ex: How many moles of O2 are required to produce 10. moles of CO2?2 CO + O2 2 CO2
1 mol CO = 2 mol H2 = 1 mol CH3OH
10. mol CO2 x __________ mol CO2
mol O2
2
1 = 5.0 mol O2
What other relationships do we have for the mole?
• 1 mol = 6.02 x 1023 atoms / molecules / particles
• 1 mol = [molar mass] g
• 1 amu = 1.66x10-24 g
We can add these mole relationships on either end of the mole ratio:
# unit A x 1 mol A x mol B x __ unit B = # unit B
_ unit A _ mol A 1 mol B
mole relationship mole ratio mole relationship
(switch units) (switch substances) (switch units)
*A is the GIVEN substance & B is the WANTED substance
Mass A – Mole B
Ex: Calculate moles of O2 produced if 2.50 g KClO3 decomposes completely:
2 KClO3 2 KCl + 3 O2
x ______________mol O2
mol KClO3
3
2
x ________________mol KClO3
122.55
1
g KClO3
2.50 g KClO3 =
0.0306 mol O2
K 1 x 39.10 = 39.10
Cl 1 x 35.45 = 35.45
O 3 x 16.00 = 48.00
122.55 g/mol
Mass A – Mass B
Ex: Determine the mass of NaCl that decomposes to yield 355 g Cl2
2NaCl 2 Na + Cl2355 g Cl2
mol Cl2
g Cl270.90
1 x _____________ x ______________ mol Cl2
mol NaCl
1
2 x ______________ mol NaCl
g NaCl58.44
1=
585 g NaCl
Cl 2 x 35.45 = 70.90 g/mol
Na 1 x 22.99 = 22.99
Cl 1 x 35.45 = 35.45
58.44 g/mol
Mole A – Mass BEx: Calculate the number of grams of oxygen
required to react exactly with 4.30 mol of propane, C3H8, in the reaction by the following balanced equation:
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)
4.30 mol C3H8 mol O2
mol C3H81
5 x _____________ x __________ mol O2
g O2
1
32.00=
688 g O2O 2 x 16.00 = 32.00 g/mol
Learning Check
How many grams of water are needed to produce 9.23 moles of oxygen?____ Na2O2 + ____ H2O ____ NaOH + ____ O2
Background Knowledge Check
Label the reactant(s) and product(s) in the following reaction:
2 Mg + O2 2MgO
Reactant(s):
Product(s):
Mg and O2
MgO
Limiting ReactantManufacturers of cars and bicycles order parts in the same
proportion as they are used in their product. Car manufacturers order four times as many wheels as engines and bicycle manufacturers order twice as many pedals as seats. In the same manner, when chemicals are mixed together so they can undergo a reaction, they are often mixed in stoichiometric quantities – exactly the correct amounts so that all the reactants “run out” at the same time.
If the chemicals aren’t mixed to run out at the same time, one of the chemicals will limit or halt the reaction from taking place any further. The reactant that “runs out” or limits the reaction is called the limiting reactant. The reactant that still remains or is extra is called the excess reactant.
In any stoichiometric problem, where reactants are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting in order to calculate correctly the amounts of products that will be formed.
Analogy: Baking CookiesA recipe calls for 2 cups of flour for every egg. You have 5
cups of flour and 3 eggs.What is your limiting ingredient? What is your excess ingredient?
Steps for Solving Stoichiometry Problems Involving Limiting Reactants1. Write and balance the equation for the reaction, if necessary.2. For each reactant, convert grams reactant to grams product.3. Compare grams of product:
• The smaller grams of product is the theoretical yield and is the amount of product made
• The smaller grams of product came from the limiting reactant• The larger grams of product came from the excess reactant
flour
eggs
x ______________ mol Mg
mol MgO2
2
x _____________mol O2
mol MgO 2
1
7.24 mol Mg=
=3.86 mol O2
292 g MgO
Ex: 7.24 moles of Mg and 3.86 moles of O2 react to form MgO. 2 Mg + O2 2MgO How many grams of MgO are formed ?
What is the limiting reactant ? What is the excess reactant ? Mg O2
Mg 1 x 24.31 = 24.31
O 1 x 16.00 = 16.00
40.31 g/mol
x ____________ mol MgO
g MgO40.31
1
x ____________ mol MgO
g MgO40.31
1311 g MgO
x ___________ g N2
mol N21
28.02 x ___________
mol N2
mol NH3 2
1
x ___________ g H2
mol H21
2.02
x ______________ mol NH3
g NH317.04
1
x ___________ mol H2
mol NH32
3
2.50 x 104 g N2
3.04 x 104 g NH3=
=
5.00 x 103 g H2
2.81 x 104 g NH3
Ex: Suppose 2.50 x 104 g of N2 and 5.00 x 103 g of H2 are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when the reaction is run to completion. N2 + 3 H2 2 NH3
What is the limiting reactant? What is the excess reactant?
N2H2
N 2 x 14.01 = 28.02 g/mol
H 2 x 1.01 = 2.02 g/mol
N 1 x 14.01 = 14.01
H 3 x 1.01 = 3.03
17.04 g/mol
x ______________ mol NH3
g NH317.04
1
Percent Yield Theoretical yield – amount of product predicted
from the amounts of reactants used, calculated from the limiting reactant
Actual yield – amount of product actually obtained through experiment
Percent yield – comparison of actual and theoretical yield
Percent Yield = Actual yield X 100%
Theoretical yield
Example: Methanol, CH3OH, can be produced by the reaction between carbon monoxide and hydrogen. Suppose 6.85 x 104 g of CO is reacted with 8.60 x 103 g of hydrogen. CO + 2H2 CH3OH1. Calculate the theoretical yield of methanol.
2. If 3.57 x 104 g of CH3OH is actually produced, what is the percent yield of methanol?
x ___________ g CO
mol CO1
28.01 x ___________
mol CO
mol CH3OH 1
1
x ___________ g H2
mol H21
2.02 x ___________
mol H2
mol CH3OH1
2
6.85 x 104 g CO
7.83 x 104 g CH3OH=
=
8.60 x 103 g H2
6.82 x 104 g CH3OH
C 1 x 12.01 = 12.01
O 1 x 16.00 = 16.00
28.01 g/mol
H 2 x 1.01 = 2.02 g/mol C 1 x 12.01 = 12.01
H 4 x 1.01 = 4.04
O 1 x 16.00 = 16.00
32.05 g/mol
% Yield =3.57 x 104 g
6.82 x 104 gX 100% = 52.3 %
6.82 x 104 g CH3OH
x ______________ mol CH3OH
g CH3OH32.05
1
x ______________ mol CH3OH
g CH3OH32.05
1
Learning CheckLearning Check - If 1.30 grams of oxygen and 3.10 grams of iron are reacted, then what is the theoretical yield of iron (III) oxide. What is the limiting reactant? If 3.00 grams of iron (III) oxide is produced, then what is the percent yield?
Balanced equation: 4 Fe + 3 O2 2 Fe2O3
Theoretical Yield?
Limiting Reactant? Excess Reactant?
Percent Yield?