1. I. I.Stoichiometric Calculations Stoichiometry – Ch. 10

2. A. Proportional Relationships b I have 5 eggs. How many cookies can I make? 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 5 eggs5 doz. 2 eggs = 12.5 dozen cookies Ratio of eggs to cookies

3. A. Proportional Relationships b Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

4. B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass -moles  grams Molar volume -moles  liters gas Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

5. C. Stoichiometry Problems 1.How many moles of KClO 3 must decompose in order to produce 9.5 moles of oxygen gas? 9.5 mol O 2 2 mol KClO 3 3 mol O 2 = 6.3 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol 9.5 mol

6. 2.How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L C. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

7. C. Stoichiometry Problems 2. How many grams of silver will be formed from 12.0 g copper in a silver nitrate solution? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

8. 107.87 g Ag 1 mol Ag C. Stoichiometry Problems 3.How many grams of silver will be formed from 15.0 g AgNO 3 when it reacts with copper wire? 15.0g AgNO 3 1 mol AgNO 3 169.88 g AgNO 3 = 9.52 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 2 mol AgNO 3 15.0 g ? g

9. D. Limiting Reactants b Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 12.0 g 15.0 g ? g 40.7 g Ag 9.52 g Ag Can’t make both amounts! One of the reactants limits the amount of product that can be produced.

10. D. 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

11. D. Limiting Reactants b Limiting Reactant – AgNO 3 used up in a reaction determines the amount of product b Excess Reactant - Cu added to ensure that the other reactant is completely used up cheaper & easier to recycle

12. D. Limiting Reactants 1. Write a balanced equation, identify known (2 now) and unknown. 2. For each reactant, calculate the amount of product formed. 2 lines of stoichiometry!!!!! 3. Smaller answer indicates: limiting reactant amount of product produced

13. D. Limiting Reactants b 79.1 g of zinc react with 82.0 g of 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 82.0 g

14. 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 82.0 g

15. A. Limiting Reactants 22.4 L H 2 1 mol H 2 1mol HCl 36.46 g = 25.2 L H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 82.0 g 82.0g HCl

16. A. Limiting Reactants Zn: 27.1 L H 2 HCl: 25.2 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25.2 L H 2 left over zinc

17. B. Percent Yield calculated on paper measured in lab or given in the problem

18. 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 Excess

19. 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:

20. B. Percent Yield Theoretical Yield = 49.4 g KCl % Yield = 46.3 g 49.4 g  100 = 93.72% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g