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

Stoichiometry – Ch. 12. II. Stoichiometry in the Real World Limiting Reagents and % yield (p. 268-378). A. Limiting Reactants. Available Ingredients 4 slices of bread 1 jar of peanut butter 1/2 jar of jelly. Limiting Reactant bread. Excess Reactants peanut butter and jelly.

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

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

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

  3. A. Limiting Reactants • Limiting Reactant • used up in a reaction • determines the amount of product • 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 • 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  ZnCl2 + H2 ? L 79.1 g 82.0g

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

  7. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 82.0g 79.1 g Zn 1 mol Zn 65.39 g Zn 1 mol H2 1 mol Zn 22.4 L H2 1 mol H2 = 27.1 L H2

  8. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 82.0 82.0 g 1 mol HCl 36.5g HCl 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2

  9. A. Limiting Reactants left over zinc Choose smallest # Zn 27.1 L H2 HCl 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2

  10. How many grams of excess reactant remain? • Use stoichiometry to convert from L.R. (HCl) to amount of E.R. (Zinc) USED..getting… • 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 measured in lab calculated on paper

  12. B. Percent Yield • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  13. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138.21 g K2CO3 2 mol KCl 1 mol K2CO3 74.55 g KCl 1 mol KCl = 49.4 g KCl

  14. B. Percent Yield 46.3 g 49.4 g K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.4 g KCl 45.8 g 49.4 g actual: 46.3 g  100 = 93.7% % Yield =

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