Question

If the yield of potassium benzoate is 68% according to the following reaction:

C₇H₈ + 2KMnO₄ ====> KC₇H₅O₂ + 2MnO₂ + KOH +H₂O

What is the minimum number of grams of toluene needed to achieve this yield and considering the actual yield of KC₇H₅O₂ is 10.0 g.

Answer 1

Given that yield of potassium benzoate is 68% the actual yield of KC7H5O2 is 10.0 g according to the following reaction СН; + 2KMnO > КС-Н:0 + 2MnOz + KOH +Н,0 Calculate the theoretical yield: % yield actual yield theoretical yield *100 68%= actual yield of KC^H5O2 is 10.0 g/ theoretical yield*100 theoretical yield 10/68*100 = 14.71 g KC7H5O2 Number of moles = amount in g/ molar mass - 14.71 g KC7H5O2/160.2117 g/mol = 0.0918 moles g KC^H5O2 Calculate the moles of toluene: 0.0918 moles g KC7H5O2 1 mole toluene /1 moles g KC7H5O2 =0.0918 moles toluene Amount in g= number of mole s* molar mass = 0.0918 moles toluene * 92.13842 g/mol = 8.45978 g toluene = 8.46 g The % yield is the observed experimental yield divided by the theoretical yield and multiplied by 100 The percentage yield cam be calculates by the use of following expression: Actual yield or given yield Theoretical yield or calculated yield x100% Theoretical yield is the greatest amount of a product possible from a reaction, when assume that the reaction goes to 100% competition.