2. The Study of Stoichiometry I. Stoichiometric Calculations

3. A. Proportional Relationships b How many eggs are needed to make 12 dozen cookies? 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 12 doz.2 eggs 5 doz. = 5 eggs Ratio of eggs to cookies

4. A. Proportional Relationships b Stoichiometry calculating amounts of reactants & products using mole ratios b Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O 2  2 MgO

5. A. Proportional Relationships b Mole Ratio examples: 2 H 2 + O 2  2 H 2 O 2 mol H 2 : 1 mol O 2  2 mol H 2 O CH 4 + 2O 2  CO 2 + 2H 2 O 1 mol CH 4 : 2 mol O 2  1 mol CO 2 : 2 mol H 2 O

6. 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.

7. 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

8. C. Stoichiometry Problems b How many grams of KCl will be formed from 2.5 mol KClO 3 ? 2.5 mol KClO 3 2 mol KCl 2 mol KClO 3 = 186.3 g KCl 2KClO 3  2KCl + 3O 2 74.55 g KCl 1 mol KCl 2.5 mol? g

9. 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

10. C. Stoichiometry Problems 1. Try on your own How many grams of calcium carbonate are required to prepare 50.0 g of calcium oxide? CaCO 3  CaO + CO 2

11. C. Stoichiometry Problems

13. 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

14. II. Gas Stoichiometry Ch. 5 - Stoichiometry

15. 1 mol of a gas=22.4 L at STP A. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

16. A. Molar Volume at STP Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

17. B. Gas Stoichiometry Problem b How many grams of CaCO 3 are req’d to produce 9.00 L of CO 2 at STP? 9.00 L CO 2 1 mol CO 2 22.4 L CO 2 = 40.2 g CaCO 3 CaCO 3  CaO + CO 2 1 mol CaCO 3 1 mol CO 2 100.09 g CaCO 3 1 mol CaCO 3 ? g9.00 L

18. StoichiometryStoichiometry Limiting Reactants

19. 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

20. 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

21. A. Limiting Reactants L 2 + 3 S 2  2 LS 3 b Obtain a bag containing: 4 L 2 - large paper clip molecules 4 S 2 - small paper clip molecules b Remove the contents of the bag. b Under the balanced equation, label how many molecules you have of each reactant. b Form as many LS 3 molecules as possible. b How many LS 3 molecules did you form? b Which reactant ran out?

22. 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

23. A. Limiting Reactants b 79.1 g of zinc react with 2.1 mol 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 2.1 mol

24. 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 2.1 mol

25. A. Limiting Reactants 22.4 L H 2 1 mol H 2 2.1 mol HCl 1 mol H 2 2 mol HCl = 23.52 L H 2 Zn + 2HCl  ZnCl 2 + H 2 79.1 g ? L 2.1 mol

26. A. Limiting Reactants Zn: 27.1 L H 2 HCl: 23.5 L H 2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 23.52 L H 2 left over zinc

27. A. Limiting Reactants b 35.6 g of zinc react with 1.9 mol HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl 2 + H 2 35.6 g ? L 1.9 mol

28. A. Limiting Reactants 35.6 g Zn 1 mol Zn 65.39 g Zn = 12.2 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 35.6 g ? L 1.9 mol

29. A. Limiting Reactants 22.4 L H 2 1 mol H 2 1.9 mol HCl 1 mol H 2 2 mol HCl = 21.28 L H 2 Zn + 2HCl  ZnCl 2 + H 2 35.6 g ? L 1.9 mol

30. A. Limiting Reactants Zn: 12.2 L H 2 HCl: 21.28 L H 2 Limiting reactant: Zn Excess reactant: HCl Product Formed: 12.2 L H 2 left over HCl

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

32. 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

33. 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:

34. 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

35.  2 HCl + Zn  H 2 + ZnCl 2 130.7 g Zn was able to produce twice its mass of Zinc Chloride in the chemical reaction above. What is the percent yield of ZnCl 2 ? = 130.7 1 65.39 g Zn Mol Zn g Zn 272.4 g ZnCl 2 g ZnCl 2 136.28 mol ZnCl 2 1 Percent Yield 1 mol Zn mol ZnCl 2 1 = Actual Calculated 261.4 g ZnCl 2 272.4 g ZnCl 2 X 100 = 95.9% Yield

36.  NaCl + H HCl + NaOH  NaCl + H 2 O 100 g HCl reacts with 100 g NaOH in the chemical reaction above. What is the limiting reactant and how many grams of NaCl will be produced by the reaction? = 100 1 36.45 g HCl mol HCl g HCl 160.3 g NaCl g NaCl 58.44 mol NaCl 1 = 100 1 40.0 g NaOH mol NaOH g NaOH 146.1 g NaCl mol NaCl 1 mol NaOH 1 mol NaCl 1 g NaCl 58.44 Limiting Reactant 1 mol HCl mol NaCl 1