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Stoichiometry Calculations Limiting Reactants and Percent Yield

Granada Hills Charter High School Student Review and Exam Preparation. Stoichiometry Calculations Limiting Reactants and Percent Yield. The Mole-ratio concepts. Mole-Mole Calculations. Mole-Mass Calculations. Mass-Mass Calculations. Percent Yield Calculations.

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Stoichiometry Calculations Limiting Reactants and Percent Yield

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  1. Granada Hills Charter High SchoolStudent Review and Exam Preparation Stoichiometry CalculationsLimiting Reactants and Percent Yield

  2. The Mole-ratio concepts Mole-Mole Calculations Mole-Mass Calculations Mass-Mass Calculations Percent Yield Calculations

  3. 2 mol KMnO4 2 mol KCl KMnO4 + HCl + H2S KCl + MnCl2+ S + H2O 2 mol KMnO4 2 mol KMnO4 2 mol KMnO4 8 mol H2O 5 mol S 5 mol H2S 5 mole S 2 mole KMnO4 Mole-Ratio Concept KMnO4 S 2 6 5 2 2 5 8 How many moles of S can be obtained from 1.5 mole KMnO4? 1.5 mole KMnO4 x = 3.8 Mole S

  4. KMnO4 + HCl + H2S KCl + MnCl2+ S+ H2O 1 mol KMnO4 5 mol S 32.07 g S 158.04 g KMnO4 2mol KMnO4 1mol S Mass-Mole-Mass Calculation 2 6 5 2 2 5 8 How many grams of S can be obtained from 1.5 g KMnO4 ? 1.5 g KMnO4 x x x = 80.9 g S

  5. Which is the limiting reactant? Strategy: • Use the relationships from the balanced chemical equation • You take each reactant in turn and ask how much product would be obtain, if each were totally consumed. • The reactant that gives the smaller amount of product is the limiting reactant.

  6. 1 mol Fe 1 mol Fe3O4 55.85 g Fe 3 mol Fe 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 1 mol H2O 1 mol Fe3O4 18.02 g H2O 4 mol H2O Limiting Reactant Calculation 16.8 g 10.0 g 1. Calculate amount of product formed by each reactant .100 mol Fe3O4 .100 mol Fe3O4 16.8 g Fe 16.8 g Fe x x = Maximum yieldof product .139 mol Fe3O4 10.0 g H2O x x = 2. The Limiting reactant gives the least amount of product.

  7. How much Excess Reactant Remains? Strategy: • Start with the original amount of the already identified Limiting Reactant. • Use the relationships from the balanced chemical equationto find how much of the excess reactant was used by the limiting reactant to make the product. • Subtract this quantity from the original amount of the excess reactant provided.

  8. L/R XS 16.8 g 10.0 g 1 mol Fe 55.85 g Fe 4 mol H2O 18.02 g H2O 16.8 g Fe x x x = 7.23gH2O 3 mol Fe 1 mol H2O 3Fe(s) + 4H2O(g) Fe3O4 + 4H2 This amount was used to make the product Excess Reactant Calculation 3. Calculate the amount of the other reactant (XS) that wasutilized by the limiting reactant to form the product. 4. Calculate the excess amount by subtracting the reacted amount from the original starting quantity Excess water: 10.0 g - 7.23 g = 2.7 g unreacted water

  9. Actual Yield x 100 Theoretical Yield 231.55 g Fe3O4 1 mol Fe3O4 16.2 g Fe3O4 23.2 g Fe3O4 Percent Yield Calculations In the previous reaction the theoretical yield was 23.2 g Fe3O4 0.100 mol Fe3O4 = x If the actual amount obtained is 16.2 g, then the % yield: Percentage Yield = x 100 = 69.8%

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