What mass of ethyl acetate can be obtained in the reaction between 1.61 g of ethanol and 1.8 g

What mass of ethyl acetate can be obtained in the reaction between 1.61 g of ethanol and 1.8 g of acetic acid if the ester yield is 75%?

Let’s find the amount of acid substance by the formula:

n = m: M.

M (CH3COOH) = 60 g / mol.

n = 1.8 g: 60 g / mol = 0.03 mol.

Let’s find the amount of ethanol substance:

M (C2H5OH) = 46 g / mol.

n = 1.61 g: 46 g / mol = 0.035 mol.

Acetic acid is deficient. We calculate the amount of ether substance according to the lack.

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

С2Н5ОН + CH3 – COОH → CH3 – COО – С2Н5 + H2O.

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

The amount of substance will be the same.

n (CH3COOH) = n (CH3 – CO O – C2H5) = 0.03 mol.

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

m = n × M,

M (CH3 – CO O – C2H5) = 88 g / mol.

m = 0.03 mol × 88 g / mol = 2.64 g.

2.55 g was calculated (theoretical yield).

According to the condition, the ether yield was 75%.

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

2.64 g – 100%,

x g – 75%,

x = (2.64 × 75%): 100% = 1.98 g.

Answer: 1.98 g.