The interaction of C2H5OH with CH3COOH gave ether with a mass fraction of 85%

The interaction of C2H5OH with CH3COOH gave ether with a mass fraction of 85% of the theoretical. How much 98% acetic acid solution is needed to obtain 220 g of ether?

С2Н5ОН + СН3СООН = С2Н5СО (О) СН3 + Н2О

Let’s find the theoretical mass of the ether:

mtheor (C2H5CO (O) CH3) = mpract (C2H5CO (O) CH3) / η = 220 / 0.85 = 258.82 g;

Let’s find the theoretical amount of the ether substance:

ntheor = mtheor (C2H5CO (O) CH3) / M (C2H5CO (O) CH3) = 258.82 / 88 = 2.94 mol;

By stoichiometry of the reaction:

n (CH3COOH) = ntheor (C2H5CO (O) CH3) = 2.94 mol;

m (CH3COOH) = n (CH3COOH) * M (CH3COOH) = 2.94 * 60 = 176.4 g;

m solution (CH3COOH) = m (CH3COOH) / w (CH3COOH) = 176.4 / 0.98 = 180 g.

Answer: m solution (CH3COOH) = 180 g.