1. Percentage Yield

2. OUTCOME QUESTION(S): C11-3-15 STOICHIOMETRY Vocabulary & Concepts
Solve stoichiometric problems involving moles, mass, and volume, given the reactants and products in a balanced chemical reaction.
Include: heat of reaction
Identify the limiting reactant and calculate the mass of a product, given reaction equation and reactant data.
Include: theoretical yield, experimental yield
Vocabulary & Concepts 
Actual Yield

3. - 5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g)?
HAVE g L -
USE
58.9 g
191 L
16.1 g (excess)
C3H8 - excess reactant O2 - limiting reactant
150 L O2
1 mol O2
3 mol CO2
22.4 L CO2
= 90 L CO2
22.4 L O2
5 mol O2
1 mol CO2
Our calculations suggest the reaction will use ALL 150 L of O2 and only 58.9 g of C3H8 to make 90 L of CO2
BUT…will it actually produce 90 L when the experiment is run?

4. Pressure (of gas reactants) Human error
Reactions are influenced by external factors which effect the amount of yield produced:
Concentration
State (s, l, g, aq)
Temperature
Pressure (of gas reactants)
Human error
Nothing is 100% efficient
Poor recovery /technique
Impurities in reactants
Miscalculation
Unexpected side reaction

5. actual yield x 100 percent yield = theoretical yield
Theoretical yield: amounts of product calculated from the limiting reagent.
Actual yield: (also experimental yield), amount produced during the conducted experiment.
Percentage yield: yield ratio (actual to theoretical) expressed as a percentage.
actual yield
theoretical yield
percent yield =
x 100

6. 5 3 4 C3H8 (g) + O2 (g) CO2 (g) + H2O (g)?
HAVE g L
USE
58.9 g
191 L
150 L O2
1 mol O2
3 CO2
22.4 L CO2
= 90 L CO2
22.4 L O2
5 O2
1 mol CO2
Theoretical Yield
An experiment is run producing 74 L of CO2 gas.
What is the % yield?
Actual Yield
actual yield
74 L
percent yield =
x 100
= 82 %
theoretical yield
90 L

7. 5. 00 g of KClO3 is heated and decomposes to yield 1. 78 g of O2
5.00 g of KClO3 is heated and decomposes to yield 1.78 g of O2. What is the % yield?
Actual Yield
2 KClO3  2 KCl + 3 O2
5.00 g
? g
mol
mol
5.00 g KClO3
1 mole
3 O2
32.0 g O2
122.5 g
2 KClO3
1 mole O2
= g O2
Theoretical Yield
actual yield
1.78 g
x 100
= 90.8 %
percent yield =
theoretical yield
1.96 g

8. Be careful rearranging this equation – don’t forget about the 100%...
2 KClO3  2 KCl + 3 O2
How much O2 would be produced if the percentage
yield was 78.5%?
Ac. yield
1.96 g
78.5 % =
% yield =
x 100
Th. yield
1.53 g
= Actual yield
Be careful rearranging this equation – don’t forget about the 100%...

9. O2 - excess reactant H2 - limiting reactant
What is the % yield if 58 g H2O are produced by
combining 60.0 g O2 and 7.0 g H2?
O2 (g) H2 (g) H2O (g)
HAVE g g
USE g
60.0 g O2
1 mole O2
2 H2
2.0 g H2
= 7.5 g H2
32.0 g O2
1 O2
1 mole H2
O2 - excess reactant H2 - limiting reactant
This is a Limiting Reactant question, BUT it doesn’t ask to find excess – so just “pick one” way and identify the LR

10. O2 (g) + 2 H2 (g) 2 H2O (g) = 63 g H2O 58 g percent yield = x 100?
HAVE g g
USE g
7.0 g H2
1 mole H2
2 H2O
18.0 g H2O
2.0 g H2
2 H2
1 mole H2O
= 63 g H2O
Actual Yield: 58 g
Theoretical Yield
58 g
actual yield
percent yield =
x 100
= 92 %
theoretical yield
63 g

11. CAN YOU / HAVE YOU? C11-3-15 STOICHIOMETRY Vocabulary & Concepts
Solve stoichiometric problems involving moles, mass, and volume, given the reactants and products in a balanced chemical reaction.
Include: heat of reaction
Identify the limiting reactant and calculate the mass of a product, given reaction equation and reactant data.
Include: theoretical yield, experimental yield
Vocabulary & Concepts 
Actual Yield