Reaction Yield.

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Limiting Reactants and Percent Yield

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Presentation transcript:

Reaction Yield

Theoretical Yield The maximum amount of product that can be produced from a given amount of reactant Theoretical yield is calculated using stoichiometry

Actual Yield The amount of product actually produced when the chemical reaction is carried out in an experiment

Percent Yield A comparison of the actual and theoretical yield In general, the higher the yield, the better the results are from the experiment. % Yield = actual yield (experiment) theoretical yield (calculation) × 100

Steps Identify what is given in the problem. One product and one reactant (go to step 2) One product and two reactants (go to step 3) The product given is the actual yield, calculate the theoretical yield The product given is the actual yield, calculate the theoretical yield from LR Calculate the percent yield

Example 1 Determine the theoretical yield of silver chromate if 0.500 g of silver nitrate is used to react with potassium chromate. If 0.455 g of silver chromate is obtained from an experiment, calculate the percent yield. 2 AgNO3 + 1 K2CrO4 1 Ag2CrO4 + 2 KNO3 0.500g AgNO3 1 mol AgNO3 1 mol AgCrO4 331.74 g AgCrO4 169.88 g AgNO3 2 mol AgNO3 1 mol AgCrO4 = 0.488 g AgCrO4 % Yield = Actual 0.455 g AgCrO4 Theoretical 0.488 g AgCrO4 ×100 = ×100 = 93.2%

Example 2 Carbon tetrachloride was prepared by reacting 100.0 g of carbon disulfide and 100.0 g of chlorine gas. Calculate the theoretical and percent yield if 65.0 g of carbon tetrachloride was obtained from the reaction. 1 CS2 + 3 Cl2 1 CCl4 + 1 S2Cl2 100.0g CS2 1 mol CS2 1 mol CCl4 153.8 g CCl4 76.15 g CS2 1 mol CS2 1 mol CCl4 = 202.0 g CCl4 100.0g Cl2 1 mol Cl2 1 mol CCl4 153.8 g Ccl4 70.90 g Cl2 3 mol Cl2 1 mol CCl4 = 72.31 g CCl4 L.R. = theoretical % Yield = Actual 65.0 g CCl4 Theoretical 72.31 g CCl4 ×100 = ×100 = 89.9%

Example 3 Silver bromide was prepared by reacting 200.0 g of magnesium bromide and 100.0 g of silver nitrate. Calculate the theoretical and percent yield if 100.0 g of silver bromide was obtained from the reaction. 1 MgBr2 + 2 AgNO3 1 Mg(NO3)2 + 2 AgBr 200.0g MgBr2 1 mol MgBr2 2 mol AgBr 187.77 g AgBr 184.11 g MgBr2 1 mol MgBr2 1 mol AgBr =408.0g AgBr = 110.5 g AgBr 100.0g AgNO3 1 mol AgNO3 2 mol AgBr 187.77 g AgBr 169.88 g AgNO3 2 mol AgNO3 1 mol AgBr L.R. = theoretical % Yield = Actual 100.0 g AgBr Theoretical 110.5 g AgBr ×100 = ×100 = 90.5%