What mass of pure ethanol must be taken to obtain 37.4 g of ethyl acetate, if the mass fraction of the ether yield is 85%?

1. Let’s write down the reaction equation:

CH3COOH + C2H5OH = CH3COO-C2H5 + H2O.

2. Find the amount of ether:

nprak (CH3COO-C2H5) = m (CH3COO-C2H5) / M (CH3COO-C2H5) = 37.4 g / 88 g / mol = 0.425 mol.

3. According to the reaction equation, we find the amount and then the mass of alcohol:

ntheor (CH3COO-C2H5) = nprak (CH3COO-C2H5) / η (CH3COO-C2H5) = 0.425 mol / 0.85 = 0.5 mol.

n (C2H5OH) = ntheor (CH3COO-C2H5) = 0.5 mol.

m (C2H5OH) = n (C2H5OH) * M (C2H5OH) = 0.5 mol * 46 g / mol = 23 g.

Answer: m (C2H5OH) = 23 g.