# Determine the mass of the salt formed by the interaction of excess sodium hydroxide and 400 g of a 6% phenol solution.

1.Let’s find the mass of dissolved phenol by the formula:

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

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

m (C6 H5OH) = (400 g × 6%): 100% = 24 g.

or:

400 g – 100%,

m (C6 H5OH) – 6%,

m (C6 H5OH) = (6% × 400g): 100% = 24g.

2.Let’s find the amount of phenol substance by the formula:

n = m: M, M (C6 H5OH) = 12 × 6 + 1 × 6 + 16 = 94 g / mol.

n = 24 g: 94 g / mol = 0.255 mol.

3. Let’s compose the equation of the reaction between sodium hydroxide and phenol, we will find the quantitative ratios of these substances.

C6H5OH + NaOH = C6H5ONa + H2O.

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

n (C6 H5 OH) = n (C6 H5 O Na) = 0.255 mol.

4. Let’s find the mass of sodium phenolate.

m = M × n,

M (C6 H5O Na) = 12 × 6 + 5 + 16 + 23 = 116g / mol.

m = 116g / mol × 0.255 mol = 29.58 g.

Answer: m (salt) = 29.58 g.