# Find the mass of 90% acetic acid solution required to obtain 11 g of ethyl acetate.

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

n = m: M.

M (CH3COOC2H5) = 88 g / mol.

n = 11 g: 88 g / mol = 0.125 mol.

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

С2Н5ОН + CH3COОH → CH3COОС2Н5 + H2O.

According to the reaction equation for 1 mol of acid there is 1 mol of ethyl acetate. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (CH3COOH) = n (CH3 CO O C2H5) = 0.125 mol.

Let’s find the mass of CH3COOH by the formula:

m = n × M,

M (CH3COOH) = 60 g / mol.

m = 0.125 mol × 60 g / mol = 7.5 g.

Find the mass of the CH3COOH solution.

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

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

m (solution) = (7.5 g: 90%) × 100% = 8.33 g.

Answer: 8.33 g.