What mass of potassium phenolate can be obtained from phenol and 120 g of KOH

What mass of potassium phenolate can be obtained from phenol and 120 g of KOH solution with a mass fraction of alkali of 14%?

1.Let’s find the mass of potassium hydroxide.

The mass of dissolved potassium hydroxide can be found by the formula:

w = m (substance): m (solution) x 100%,

hence m (substance) = (m (solution) × w): 100%.

m (KOH) = (120 g × 14%): 100% = 16.8 g.

Or:

120 g – 100%,

X g – 14%,

X = (120 × 14%): 100%,

X = 16.8 g.

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

n = m: M,

M (KOH) = 39 + 1 + 16 = 56 g / mol.

n = 16.8 g: 56 g / mol = 0.3 mol.

We compose the reaction equation. We find in what quantitative ratios the substances are.

C6 H5OH + KOH = C6H5OK + H2O

1 mol of potassium hydroxide reacts with 1 mol of phenol, 1 mol of potassium phenolate and 1 mol of water are formed. The ratio of substances is 1: 1. Then the amount of phenol and potassium hydroxide will be the same.

n (C6H5OK) = 0.3 mol.

3.Let’s find the mass of potassium phenolate by the formula:

m = M × n,

M (C6H5OK) = 12 × 6 + 5 + 16 + 39 = 132 g / mol.

m = 132 g / mol × 0.3 mol = 39.6 g.

Answer: 39.6 g.