Calculate the mass of the ether that is formed by the interaction of absolute ethyl alcohol with a mass of 18.4
Calculate the mass of the ether that is formed by the interaction of absolute ethyl alcohol with a mass of 18.4 with glacial acetic acid with a mass of 18, take the ether yield equal to 70 of the theoretically possible.
1.Let’s find the amount of acetic acid substance by the formula:
n = m: M.
M (CH3COOH) = 60 g / mol.
n = 18 g: 60 g / mol = 0.3 mol.
2.Let’s find the amount of the substance of ethyl alcohol.
M (C2H5OH) = 46 g / mol.
n = 18.4 g: 46 g / mol = 0.4 mol.
Acetic acid is deficient. We solve the problem by lack (by acid).
3. 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) = 0.3 mol.
4.Let’s find the mass of ethyl acetate by the formula:
m = n × M,
M (CH3 – COO – C2H5) = 88 g / mol.
m = 0.3 mol × 88 g / mol = 26.4 g.
5. 26.4 g were calculated (theoretical yield).
According to the condition of the problem, the ether received 70%.
Let’s find a practical way out of the product.
26.4 g – 100%
x – 70%,
x = (26.4 × 70%): 100 g = 18.48 g.
Answer: 18.48 g.