# To obtain ethyl acetate, 91 g of alcohol with a mass fraction of 96% and 122, 45 g of an acid solution with a mass fraction

**To obtain ethyl acetate, 91 g of alcohol with a mass fraction of 96% and 122, 45 g of an acid solution with a mass fraction of 98% were taken. Determine the mass of the ester obtained, considering that the reaction yield is practically 96%.**

W = m (substance): m (solution) × 100%,

m (substance) = (m (solution) × W): 100%.

m (C2H5OH) = (91 g × 96%): 100% = 87.36 g.

m (CH3COOH) = (122.45 g × 98%): 100% = 120 g.

Let’s find the amount of substance С2Н5ОН according to the formula:

n = m: M.

M (C2H5OH) = 46 g / mol.

n = 87.36 g: 46 g / mol = 1.9 mol.

M (CH3COOH) = 60 g / mol.

n = 120 g: 60 g / mol = 2 mol (excess).

Let’s compose the reaction equation, find the quantitative ratios of substances.

С2Н5ОН + CH3COОH → CH3COОС2Н5 + H2O.

According to the reaction equation, 1 mol of C2H5OH accounts for 1 mol of ethyl acetate. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (C2H5OH) = n (CH3 COOC2H5) = 1.9 mol.

Let’s find the mass of ethyl acetate by the formula:

m = n × M,

M (CH3COOC2H5) = 88 g / mol.

m = 1.9 mol × 88 g / mol = 167.2 g.

167.2 g were calculated (theoretical yield).

According to the condition of the problem, the aldehyde was 96%.

Let’s find a practical way out of the product.

167.2 g – 100%,

x g – 96%,

x = (167.2 g × 96%): 100% = 160.51 g.

Answer: 160.51 g.