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Stoichiometry – Ch 9

Stoichiometry – Ch 9. Limiting/Excess Reactants a nd Percent Yield. Limiting Reactants. Available Ingredients 4 slices of bread 1 jar of peanut butter ½ jar of jelly Limiting Reactant: bread Excess Reactants: peanut butter and jelly. Limiting Reactants. Limiting Reactant

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Stoichiometry – Ch 9

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  1. Stoichiometry – Ch 9 Limiting/Excess Reactants and Percent Yield

  2. Limiting Reactants • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • ½ jar of jelly • Limiting Reactant: bread • Excess Reactants: peanut butter and jelly

  3. 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 and easier to recycle

  4. Limiting Reactants • Write a balanced equation. • For each reactant, calculate the amount of product formed. • Smaller answer indicates: • Limiting Reactant • Amount of product

  5. Limiting Reactants 79.1 g of zinc react with 2.25 moles of HCl. Indentify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 79.1 g 2.25 mole ? L

  6. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 2.25 mole ? L 79.1 g Zn x 1mol Zn x1 mol H2x22.4 L H2 1 65 g Zn 1 mol Zn 1 mol H2 = 27.3 L H2

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

  8. Limiting Reactants Starting with Zn – 27.3 L H2 Starting with HCl – 25.2 L H2 Limiting Reactant: HCl Excess Reactant: Zn Product Formed: 25.2 L H2

  9. Percent Yield % yield = actual yield (measure in lab) x 100 theoretical yield (calculate)

  10. Percent Yield When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and percent yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g 46.3 g (actual yield)

  11. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g (theoretical yield) 46.3 g (actual yield) Theoretical yield: 45.8 g K2CO3 x 1mol K2CO3x 2 mol KCl x 74.5 g KCl 1 138 g K2CO3 1 mol K2CO3 1 mol KCl = 49. 5 g KCl

  12. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.5 g (theoretical yield) 46.3 g (actual yield) % yield = 46.3 x 100 = 93.5% 49.5

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