Answer:
9.359 g of ethyl butyrate
% yield = 58.766 %
Explanation:
The reaction is:
C4H8O2 + C2H5OH -> C6H12O2 + H2O
Molecular weight of butanoic acid: 88 g/mole
Molecular weight of ethyl butyrate: 116 g/mole
Assuming 100% yield, all 7.10 g of butanoic acid reacts. In moles are:
(7.10 g) / (88 g/mole) = 0.0806 moles of butanoic acid
From the balanced equation we know that 1 mole of butanoic acid produce 1 mole of ethyl butyrate, then 0.0806 moles of ethyl butyrate are produced. In grams are:
0.0806 moles * 116 g/mole = 9.359 g of ethyl butyrate (this represents the theoretical yield)
If 5.50 g of ethyl butyrate are produced, then the percent yield is:
% yield = (actual yield /theoretical yield) * 100
% yield = (5.50 g/9.359 g) * 100
% yield = 58.766 %
