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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 • The cheaper and more common chemical is used in excess

  4. A. Limiting Reactants 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. (this means you will set up 2 stoich problems) 3. Smaller answer indicates: • limiting reactant • amount of product that can be made

  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 Theoretical Yield: 25 L H2

  9. B. Percent Yield measured in lab • calculated on paper • Determined by limiting reactant

  10. B. Percent Yield • When 45.8 g of K2CO3 was reacted with excess HCl, 46.3 g of KCl was formed. a)Calculate the theoretical yield of KCl b)Calculte the % yield K2CO3 + 2HCl  2KCl + H2O + CO2 ? g 45.8 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 =

  13. Finding the Amount of Excess Reactant Left Over • calculate the amount of the excess reactant used • Set up another stoich problem. • Start with the LIMITING REACTANT and solve for the EXCESS REACTANT • subtract the amount of excess reactant used from the amount given to find the amount left over. • Can we find the amount of excess reactant in the next problem?

  14. 4Na (s)+ O2(g) 2 Na2O(s) • 5.00 g of sodium reacted with 5.00 g of oxygen a. How many grams of product can form? b. What is the limiting reactant? c. How much excess reactant is left over? (hint: first find the amount of excess reactant used in the reaction, and subtract from the amount given)

  15. 6.74 g sodium oxide • Na • 3.26 g of oxygen was left over

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