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Mastering Stoichiometry in Chemistry

Understand limiting reactants, available ingredients, and excess reactants in chemical reactions. Learn how to identify and calculate the limiting reactant and determine the amount of product formed. Practice with example problems for a hands-on approach to stoichiometry.

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Mastering Stoichiometry in Chemistry

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  1. Stoichiometry – Ch. 9 II. Stoichiometry in the Real World(p. 288-294)

  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 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 0.90 L 2.5M

  6. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 0.90 L 2.5M 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 0.90 L 2.5M 0.90 L 2.5 mol HCl 1 L 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. B. Percent Yield measured in lab calculated on paper

  10. 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

  11. 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

  12. 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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