How many grams of ester can be obtained from 200 g of 10% methyl alcohol and 200 g of 60%

| July 24, 2021 | education

How many grams of ester can be obtained from 200 g of 10% methyl alcohol and 200 g of 60% butyric acid if the product yield is 50%?

1.Let’s find the mass of CH3OH in the solution.

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

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

m (substance) = (200 g × 10%): 100% = 20 g.

2.Let’s find the amount of CH3OH substance.

M (CH3OH) = 32 g / mol.

n = 20 g: 32 g / mol = 0.625 mol.

Let’s find the mass of Н3ССН2 СН2 СООН in solution.

m (substance) = (200 g × 60%): 100% = 120 g.

M (H3CCH2 CH2 COOH) = 88 g / mol.

n = 120 g: 88 g / mol = 1.36 mol (excess).

Let’s find the quantitative ratios of substances.

Н3ССН2 СН2 СООН + СН3ОН → Н3ССН2 СН2 СООСН3 + Н2О.

For 1 mol of CH3OH, there is 1 mol of H3CCH2 CH2 COOCH3.

Substances are in quantitative ratios 1: 1.

n (Н3ССН2 СН2 СООСН3) = n (СН3ОН) = 0.625 mol.

3.Let’s find the mass Н3ССН2 СН2 СООСН3.

m = nM.

M (Н3ССН2 СН2 СООСН3) = 102 g / mol.

m = 0.625 mol × 102 g / mol = 63.75 g.

63.75 g – 100% (in theory),

X – 50%,

X = (63.75 × 50%): 100% = 31.875 g.

Answer: 31.875 g.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.