1. Limiting/Excess Reactants and Percent Yield
Stoichiometry – Ch 9
Limiting/Excess Reactants
and
Percent Yield

2. Limiting Reactants Available Ingredients Limiting Reactant: bread
4 slices of bread
1 jar of peanut butter
½ jar of jelly
Limiting Reactant: bread
Excess Reactants: peanut butter and jelly

3. Limiting Reactants Limiting Reactant Excess Reactant
Used up in a reaction
Determines the amount of product
Excess Reactant
Added to ensure that the other reactant is completely used up
Cheaper and easier to recycle

4. Limiting Reactants Write a balanced equation.
For each reactant, calculate the amount of product formed.
Smaller answer indicates:
Limiting Reactant
Amount of product

5. Limiting Reactants
79.1 g of zinc react with 2.25 moles of HCl. Indentify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H g 2.25 mole ? L

6. Limiting Reactants
Zn + 2HCl  ZnCl2 + H g 2.25 mole ? L 79.1 g Zn x 1mol Zn x 1 mol H2x 22.4 L H g Zn 1 mol Zn 1 mol H2 = 27.3 L H2

7. Limiting Reactants
Zn + 2HCl  ZnCl2 + H g 2.25 mole ? L 2.25 mol HCl x 1 mol H2 x 22.4 L H2 1 2 mol HCl 1 mol H2 = L H2

8. Limiting Reactants Limiting Reactant: HCl Excess Reactant: Zn
Starting with Zn – 27.3 L H2
Starting with HCl – 25.2 L H2
Limiting Reactant: HCl
Excess Reactant: Zn
Product Formed: 25.2 L H2

9. Percent Yield
% yield = actual yield (measure in lab) x 100 theoretical yield (calculate)

10. Percent Yield
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and percent yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO g ? g 46.3 g (actual yield)

11. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g ? g (theoretical yield)
46.3 g (actual yield)
Theoretical yield:
45.8 g K2CO3 x 1mol K2CO3 x 2 mol KCl x 74.5 g KCl
g K2CO mol K2CO mol KCl
= g KCl

12. Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO g 49.5 g (theoretical yield) 46.3 g (actual yield) % yield = 46.3 x 100 = 93.5% 49.5