Question

Given 7.55 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer 1

mC3H7COOC2H5: (12×6)+(1×12)+(16×2) = 116 g/mol
mC3H7COOH: (12×4)+(1×8)+(16×2) = 88 g/mol

88g....................................................116g
C3H7COOH + C2H5OH ---> C3H7COOC2H5 + H2O
7,55g..................................................X

X = (116×7,55)/88
X = 9,952g of ethyl butyrate