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II. Stoichiometry in the Real World

Stoichiometry. II. Stoichiometry in the Real World. 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. A. Limiting Reactants. Limiting Reactant used up in a reaction

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II. Stoichiometry in the Real World

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  1. Stoichiometry II. Stoichiometry in the Real World

  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 • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  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 2.25 mol of HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 2.25 mol

  6. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 2.25 mol 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

  7. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 2.25 mol 2.25 mol HCl 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 = 25 L H2

  8. A. Limiting Reactants left over zinc Zn: 27.1 L H2 HCl: 25 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25 L H2

  9. Finding the Amount of Excess • By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amountfrom the given amount to find the amount of excess. • Can we find the amount of excess potassium in the next problem?

  10. Finding Excess Practice • 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2 2 KI • Iodine is the limiting reactant, and 19.6 g of potassium iodide are produced. 15.0 g I2 1 mol I2 2 mol K 39.1 g K 254 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it! Given amount of excess reactant

  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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