Determine the mass of ethanol taken to obtain 140.8 g of ethyl acetate, provided that the ether

Determine the mass of ethanol taken to obtain 140.8 g of ethyl acetate, provided that the ether yield is 80% of theoretically possible.

C2H5OH + CH3COOH → (H +, t) CH3COOC2H5 + H2O.

Let’s compose the proportion, where x is the practical mass (CH3COOC2H5):

140.8g – 80%
X – 100%;

X = (140.8 * 100) / 80 = 176g.

M (CH3COOC2H5) = 12 + 3 + 12 + 32 + 24 + 5 = 88 g / mol.
n (CH3COOC2H5) = m / M = 176g / (88g / mol) = 2 mol.
n (CH3COOC2H5) = n (C2H5OH).
M (C2H5OH) = 24 + 5 + 16 + 1 = 46g / mol.
m (C2H5OH) = n * M = 2 mol * (46g / mol) = 92g.

Answer: m (C2H5OH) = 92g.