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Ethyl acetate is a sweet-smelling solvent used in varnishes and fingernail polish remover. It is produced industrially by heating acetic acid and ethanol
together in the presence of sulfuric acid, which is added to speed up the
reaction. The ethyl acetate is distilled off as it is formed. The equation for the
process is as follows.

CH3COOH + CH3CH2OH --> CH3COOCH2CH3 + H2O

Determine the percentage yield in the following cases:
a. 68.3 g of ethyl acetate should be produced but only 43.9 g is recovered.
b. 0.0419 mol of ethyl acetate is produced but 0.0722 mol is expected. (Hint:
Percentage yield can also be calculated by dividing the actual yield in moles
by the theoretical yield in moles.)
c. 4.29 mol of ethanol is reacted with excess acetic acid, but only 2.98 mol of
ethyl acetate is produced.
d. A mixture of 0.58 mol ethanol and 0.82 mol acetic acid is reacted and 0.46
mol ethyl acetate is produced. (Hint: What is the limiting reactant?)

The percentage yields are as follows :

a) 64.28%

b) 58.03%

c) 69.46%

d) 79.31%

Explanation:

The percentage yield is the ratio of the actual yield of a product to the expected or theoretical yield expressed as a percentage.

Percent yield = actual yield/theoretical yield x 100%

a) from the question, expected yield is 68.3 g whilst the actual yield is 43.9 g

therefore,

Percentage yield = 43.9 g/68.3 g x 100%

                             =  0.64275 x 100%

                             =  64.28%

b) From the question, expected yield is 0.0722 mol while the actual yield is 0.0419 mol. Therefore

Percentage yield = 0.0419 /0.0722 x 100%

                             = 0.58033 x 100%

                             = 58.03%

c) A limiting reactant is the reactant is is completely used in a recation to form a product. That is, it is the reactant that is not in excess.

From the equation of the reaction :

CH3COOH + CH3CH2OH --> CH3COOCH2CH3 + H2O

1 mole of ethanol reacts with 1 mole of acetic acid to give 1 mole of ethyl acetate.

From the question, ethanol is the limiting reactant because it is not in excess. Hence,

4.29 mole of ethanol should produce 4.29 mole of ethyl acetate (expected yield).

However only 2.98 mol of ethyl acetate is produced (actual yield)

Percentage yield = 2.98mol/4.29 mol x 100%

                            =  0.69463 x 100%

                            = 69.46%

d) From the question, acetic acid is in excess hence ethanol is the limiting reactant

therefore from the equation of the reaction:

1 mole of ethanol reacts with 1 mole of acetic acid to give 1 mole of ethyl acetate.

0,58 mol of ethanol should produce 0.58 mol of ethyl acetate (expected yield)

But 0.46 mol of ethyl acetae was produced (actual yield)

Percentage yield = 0.46mol /0.58 mol x 100%

                             = 0.79310 x 100%

                              = 79.31%