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Stoichiometry Continued…

Stoichiometry Continued…. II. Gas Stoichiometry. A. Molar Volume at STP. S tandard T emperature & P ressure 0°C and 1 atm. 1 mol of a gas=22.4 L at STP. A. Molar Volume at STP. LITERS OF GAS AT STP. Molar Volume ( 22.4 L/mol). MASS IN GRAMS. NUMBER OF PARTICLES. MOLES.

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Stoichiometry Continued…

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  1. Stoichiometry Continued… II. Gas Stoichiometry

  2. A. Molar Volume at STP StandardTemperature&Pressure 0°C and 1 atm 1 mol of a gas=22.4 L at STP

  3. A. Molar Volume at STP LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS NUMBER OF PARTICLES MOLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

  4. B. Gas Stoichiometry Problem • How many grams of CaCO3 are req’d to produce 9.00 L of CO2 at STP? CaCO3 CaO + CO2 ? g 9.00 L 9.00 L CO2 1 mol CO2 22.4 L CO2 1 mol CaCO3 1 mol CO2 100.09 g CaCO3 1 mol CaCO3 = 40.2 g CaCO3

  5. Stoichiometry III.Adjusting to Reality

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. B. Percent Yield measured in lab calculated on paper

  14. 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

  15. 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

  16. 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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