1. Stoichiometric Calculations Start Your Book Problems NOW!! Stoichiometry

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 Core step in all stoichiometry problems!! Mole ratio - moles  moles 4. Check answer.

5. C. Stoichiometry Problems b How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

6. b How many grams of KClO 3 are req’d to produce 9.00 mol of oxygen gas? 9.00 mol O 2 = 735 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 mol C. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

7. C. Stoichiometry Problems b How many grams of silver will be formed from 12.0 g copper? 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. 63.55 g Cu 1 mol Cu C. Stoichiometry Problems b How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M

9. II. Stoichiometry in the Real World Stoichiometry Part II

10. 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

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

12. 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

13. A. Limiting Reactants b 79.1 g of zinc react with 2.25 mol of HCl. Identify the limiting and excess reactants. How many grams of hydrogen gas are formed. Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 2.25 mol

14. A. Limiting Reactants 79.1 g Zn 1 mol Zn 65.39 g Zn = 2.44 g H 2 1 mol H 2 1 mol Zn 2.02 g H 2 1 mol H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 2.25 mol

15. A. Limiting Reactants 2.02 g L H 2 1 mol H 2 2.25 mol HCl = 2.27 g H 2 1 mol H 2 2 mol HCl Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? g 2.25 mol

16. A. Limiting Reactants Zn: 2.44 g H 2 HCl: 2.27 g H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 2.27 g H 2 left over zinc

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

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

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 = 93.7% K 2 CO 3 + 2HCl  2KCl + H 2 O + CO 2 45.8 g49.4 g actual: 46.3 g