By saponification of ethyl acetate, 34.5 g of ethanol were obtained, determine the mass of the ester reacted.

1. Let’s write down the reaction of saponification of ethyl acetate:

CH3COOC2H5 + NaOH → CH3COONa + C2H5OH;

2.Calculate the chemical amount of the ethanol formed:

n (C2H5OH) = m (C2H5OH): M (C2H5OH);

M (C2H5OH) = 2 * 12 + 5 + 16 + 1 = 46 g / mol;

n (C2H5OH) = 34.5: 46 = 0.75 mol;

3. Determine the amount and weight of the ester:

n (CH3COOC2H5) = n (C2H5OH) = 0.75 mol;

m (CH3COOC2H5) = n (CH3COOC2H5) * M (CH3COOC2H5);

M (CH3COOC2H5) = 12 + 3 + 12 + 2 * 16 + 2 * 12 +5 = 88 g / mol;

m (CH3COOC2H5) = 0.75 * 88 = 66 g.

Answer: 66 g.