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Limiting Reactants & Percent Yield

Limiting Reactants & Percent Yield. Limiting Reactants. Excess. Limiting Reactant  Reactant used up first Excess Reactant  Reactant left over when the reaction stops. Limiting. Which is the limiting factor?. 2H 2 + O 2  2H 2 O. Have: 100 mol H 2 1 mol O 2

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Limiting Reactants & Percent Yield

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  1. Limiting Reactants & Percent Yield

  2. Limiting Reactants Excess Limiting Reactant Reactant used up first Excess Reactant  Reactant left over when the reaction stops Limiting

  3. Which is the limiting factor? 2H2 + O2 2H2O Have: 100 mol H2 1 mol O2 Need: 2 mol H2 50 mol O2 O2 Limiting H2 Limiting Have: 1 mol H2 2 mol O2 Need: 4 mol H2 0.5 mol O2

  4. What is the limiting reactant when 80.0g Cu reacts with 25.0g S? Cu + S  Cu S Figure What You Have: Figure What You Need: 2 2 1 mol 63.5 g 80.0g Cu X = 1.26 mol Cu 1 mol 32.1 g = .779 mol S 25.0g S X Cu is Limiting Reactant 1 mol S 2 mol Cu = 0.630 mol S 1.26 mol Cu X

  5. Percent Yield • Yield is the amount of product produced • Theoretical Yield how much should be made – calculated • Actual Yield  how much was made - measured Actual Yield Theoretical Yield X 100% = Percent Yield

  6. What is the percent yield if 13.1g of CaO is produced when 24.8g of CaCO3 is heated? Measured CaCO3 CaO + O2 24.8g ?g Step 1: figure out molar mass 1 X 40.1 = 40.1 1 X 16.0 = 16.0 56.1 Ca O Ca C O 1 X 40.1 = 40.1 1 X 12.0 = 12.0 3 X 16.0 = 48.0 100.1

  7. Step 2: figure out how many moles you have 1 mol CaCO3 100.1g CaCO3 = .248 mol CaCO3 24.8g CaCO3 X 1 mol CaO 1 mol CaCO3 = .248 mol CaO .248 mol CaCO3 X

  8. Step 3: figure out how grams of unknown 56.1g CaO 1 mol CaO = 13.9g CaO .248 mol CaOX Theoretical Yield

  9. Step 4: figure out % Yield Actual Yield Theoretical Yield = Percent Yield X 100% 13.1g CaO 13.9g CaO = 94.2% X 100%

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