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Reaction Yield.

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Presentation on theme: "Reaction Yield."— Presentation transcript:

1 Reaction Yield

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

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

4 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

5 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

6 Example 1 Determine the theoretical yield of silver chromate if g of silver nitrate is used to react with potassium chromate. If g of silver chromate is obtained from an experiment, calculate the percent yield. 2 AgNO3 + 1 K2CrO Ag2CrO KNO3 0.500g AgNO3 1 mol AgNO3 1 mol AgCrO g AgCrO4 g AgNO mol AgNO3 1 mol AgCrO4 = g AgCrO4 % Yield = Actual g AgCrO4 Theoretical g AgCrO4 × = ×100 = 93.2%

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

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


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