Determine the mass of the ester that can be obtained by the reaction of 92 g of ethyl alcohol

Determine the mass of the ester that can be obtained by the reaction of 92 g of ethyl alcohol and 100 g of acetic acid.

Let’s execute the solution:

According to the condition of the problem, we will write the data:
CH3COOH + C2H5OH = CH3COO – C2H5 + H2O – esterification, ethyl acetate is isolated;

Calculations according to the formulas of substances:
M (CH3COOH) = 60 g / mol;

M (C2H5OH) = 46 g / mol.

Let’s determine the amount of starting materials:
Y (CH3COOH) = m / M = 100/60 = 0.16 mol (deficient substance);

Y (sl. Ether) = 0.16 mol since the amount of these substances is 1 mol.

Y (C2H5OH) = m / M = 92/46 = 2 mol (substance in excess).

Calculations are carried out for the substance in deficiency.

Find the mass of the product:
m (sl. ether) = Y * M = 0.16 * 88 = 14.08 g

Answer: ethyl acetate was obtained with a mass of 14.08 g