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IV. Stoichiometric Calculations

Stoichiometry – Ch. 3. IV. Stoichiometric Calculations. A. Proportional Relationships. Stoichiometry mass relationships between substances in a chemical reaction based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation. 2 Mg + O 2  2 MgO

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IV. Stoichiometric Calculations

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  1. Stoichiometry – Ch. 3 IV. Stoichiometric Calculations

  2. A. Proportional Relationships • Stoichiometry • mass relationships between substances in a chemical reaction • based on the mole ratio • Mole Ratio • indicated by coefficients in a balanced equation 2 Mg + O2 2 MgO 2 moles Mg + 1 mole O2 produces 2 moles MgO

  3. B. Stoichiometry Steps 1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. • Mole ratio - moles  moles • Molar mass - moles  grams • Molar volume - moles  liters gas • Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer.

  4. B. Stoichiometry Problems 1.How many moles of KClO3 must decompose in order to produce 9.5 moles of oxygen gas? 2KClO3 2KCl + 3O2 9.5 mol ? mol 9.5 mol O2 2 mol KClO3 3 mol O2 = 6.3 mol KClO3

  5. B. Stoichiometry Problems 2. How many grams of silver will be formed from 12.0 g copper in a silver nitrate solution? Cu + 2AgNO3 2Ag + Cu(NO3)2 12.0 g ? g 12.0 g Cu 1 mol Cu 63.55 g Cu 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag = 40.7 g Ag

  6. B. Stoichiometry Problems 3.How many grams of silver will be formed from 15.0 g AgNO3 when it reacts with copper wire? Cu + 2AgNO3 2Ag + Cu(NO3)2 15.0 g ? g 2 mol Ag 2 mol AgNO3 1 mol AgNO3 169.88 g AgNO3 107.87 g Ag 1 mol Ag 15.0g AgNO3 = 9.52 g Ag

  7. C. Limiting Reactants • Cu + 2AgNO3 2Ag + Cu(NO3)2 15.0 g 12.0 g ? g 40.7 g Ag 9.52 g Ag Can’t make both amounts! One of the reactants limits the amount of product that can be produced.

  8. C. 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

  9. C. Limiting Reactants • Limiting Reactant – AgNO3 • used up in a reaction • determines the amount of product • the smallest answer • Excess Reactant - Cu • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  10. C. Limiting Reactants • 79.1 g of zinc react with 82.0 g 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 82.0 g

  11. C. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 82.0 g 79.1 g Zn 1 mol Zn 65.41 g Zn 1 mol H2 1 mol Zn 22.4 L H2 1 mol H2 = 27.1 L H2

  12. C. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 ? L 79.1 g 82.0 g 1mol HCl 36.46 g 1 mol H2 2 mol HCl 22.4 L H2 1 mol H2 82.0g HCl = 25.2 L H2

  13. C. Limiting Reactants left over zinc Zn: 27.1 L H2 HCl: 25.2 L H2 Limiting reactant: HCl Excess reactant: Zn Product Formed: 25.2 L H2

  14. D. Percent Yield measured in lab calculated on paper

  15. D. 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 Excess actual: 46.3 g

  16. D. 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

  17. D. 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 = 93.72% % Yield =

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