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Understanding Percent Yield in Chemical Reactions

Learn how chemists measure the efficiency of reactions by calculating percent yield. Explore stoichiometry problems to determine theoretical and actual yields.

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Understanding Percent Yield in Chemical Reactions

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  1. Percent Yield in a Chemical Reaction

  2. Percent Yield Sometimes all the product in a chemical reaction is not produced. Some of the reactant might produce other products or some of the reactants might be spilled when transferred from the balance to the test tube or beaker.

  3. Chemists are interested in the efficiency of the reaction. The efficiency is measured by comparing the actual yield to the theoretical yield.

  4. Actual yield = the amount of product produced by actually performing the reaction in lab and measuring its mass. Theoretical yield = the amount of product that should be produced based on a stoichiometry calculation.

  5. Stoichiometry Problem Zn + 2 HCl a ZnCl2 + H2 A lab group dropped 6.5 g of zinc into a test tube containing hydrochloric acid (HCl). After the reaction was over they measured 12 g of ZnCl2 produced by this reaction. What is the theoretical yield? What is the percent yield?

  6. Stoichiometry Problem Zn + 2 HCl a ZnCl2 + H2 6.5 g 12 g What is the theoretical yield? 1 mol Zn 136 g ZnCl2 1 mol ZnCl2 6.5 g Zn x ___________ x ____________ x ____________ = 13.6 g ZnCl2 65 g Zn 1 mol ZnCl2 1 mol Zn

  7. 1 mol Zn 136 g ZnCl2 1 mol ZnCl2 6.5 g Zn x ___________ x ____________ x ____________ = 13.6 g ZnCl2 65 g Zn 1 mol ZnCl2 1 mol Zn Stoichiometry Problem Zn + 2 HCl a ZnCl2 + H2 6.5 g 12 g = actual yield 13.6 = theoretical yield

  8. Stoichiometry Problem Zn + 2 HCl a ZnCl2 + H2 6.5 g 12 g = actual yield 13.6 g = theoretical yield What is the percent yield? 12 g = 88 % % yield = X 100 13.6 g

  9. Another lab group burned 3.0 g of magnesium and collected 4.2 g of MgO. What is their percent yield? 2 Mg + O2a 2 MgO 3.0 g 4.2 g 4.2 g % yield = X 100 theoretical yield

  10. 2 Mg + O2a 2 MgO 4.2 g = actual yield 3.0 g 0.125 mole 0.125 mole 5.0 g = theoretical yield Theoretical yield: 1 mol Mg 40 g MgO 2 mol MgO 3.0 g Mg x ___________ x ____________ x ____________ = 5.0 g MgO 24 g Mg 1 mol MgO 2 mol Mg actual yield 4.2 g = 84 % % yield = X 100 = X 100 theoretical yield 5.0 g

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