What is the mass of organic matter obtained by the interaction of 50 g of acetic acid and 40 g of ethyl alcohol?

First, we write down the equation of reactions.
CH3COOH + C2H5OH -> CH3COOC2H5 + H2O.
Find the amount of acetic acid and ethyl alcohol. We write down the solution.
n = m / M.
M (CH3COOH) = 12 × 2 + 4 + 16 × 2 = 60 g / mol.
n (CH3COOH) = 50 g / 60 g / mol = 0.833 mol.
M (C2H5OH) = 12 × 2 + 6 + 16 = 46 g / mol.
n (C2H5OH) = 40 g / 46 g / mol = 0.87 mol.
Since ethyl alcohol was taken in excess, it means that we count on acetic acid.
0.83 mol CH3COOH – x mol CH3COOC2H5.
1 mol CH3COOH – 1 mol CH3COOC2H5.
x = 1 × 0.83 ÷ 1 = 0.83 mol of ethyl acetate.
n = m / M.
m = n × M.
M (CH3COOC2H5) = 8 + 12 × 4 + 32 = 88 g / mol.
m = 0.83 mol × 88 g / mol = 73.04 g.
Answer: m = 73.04 g.