1. II. Stoichiometry in the Real World Limiting Reagents and % yield (p. 268-378) Stoichiometry – Ch. 12

2. A. Limiting Reactants b Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly b Limiting Reactant bread b Excess Reactants peanut butter and jelly

3. A. Limiting Reactants b Limiting Reactant used up in a reaction determines the amount of product b Excess Reactant reactant left over… cheaper & easier to recycle, added to ensure that the other reactant is completely used up

4. A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: limiting reactant amount of product

5. A. Limiting Reactants b 79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 82.0g

6. Summary of Limiting Reactant b Do stoichiometry twice to go from each reactant to same product Now you know LR and actual amount produced b Do stoichio once more from LR to ER to determine amount used…then subtract

7. A. Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 27.1 L H 2 1 mol H 2 1 mol Zn 22.4 L H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 82.0g

8. A. Limiting Reactants 22.4 L H 2 1 mol H 2 82.0 g 1 mol HCl 36.5g HCl = 25 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 82.0

9. A. Limiting Reactants Zn  27.1 L H 2 HCl  25 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H 2 left over zinc Choose smallest #

10. How many grams of excess reactant remain? b Use stoichiometry to convert from L.R. (HCl) to amount of E.R. (Zinc) USED..getting… b Now subtract amount ER (73.5g) used from starting ER (79.1g) to get amount remaining (79.1 -73.5) = 73.5g Zn used 5.6grams of Zinc remaining

11. B. Percent Yield calculated on paper measured in lab

12. B. Percent Yield b When 45.8 g of K 2 CO 3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g

13. B. Percent Yield 45.8 g K 2 CO 3 1 mol K 2 CO 3 138.21 g K 2 CO 3 = 49.4 g KCl 2 mol KCl 1 mol K 2 CO 3 74.55 g KCl 1 mol KCl K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g? g actual: 46.3 g Theoretical Yield:

14. B. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g