To neutralize the mixture of acetic acid and phenol, 57.7 ml of potassium hydroxide solution with a mass fraction

To neutralize the mixture of acetic acid and phenol, 57.7 ml of potassium hydroxide solution with a mass fraction of 13% (density 1.12 g / ml) was required. when acting on the same mixture of bromine water, a precipitate with a mass of 33.1 g precipitated, find the masses of phenol and acetic acid in the initial mixture.

Given:
V solution (KOH) = 57.7 ml
ω (KOH) = 13%
ρ solution (KOH) = 1.12 g / ml
m (sediment) = 33.1 g

To find:
m (C6H5OH) -?
m (CH3COOH) -?

1) C6H5OH + 3Br2 => C6H2Br3OH ↓ + 3HBr;
2) n (C6H2Br3OH) = m / M = 33.1 / 331 = 0.1 mol;
3) n (C6H5OH) = n (C6H2Br3OH) = 0.1 mol;
4) m (C6H5OH) = n * M = 0.1 * 94 = 9.4 g;
5) C6H5OH + KOH => C6H5OK + H2O;
CH3COOH + KOH => CH3COOK + H2O;
6) m solution (KOH) = ρ solution * V solution = 1.12 * 57.7 = 64.624 g;
7) m (KOH) = ω * m solution / 100% = 13% * 64.624 / 100% = 8.401 g;
8) n total (KOH) = m / M = 8.401 / 56 = 0.15 mol;
9) n1 (KOH) = n (C6H5OH) = 0.1 mol;
10) n2 (KOH) = n total. (KOH) – n1 (KOH) = 0.15 – 0.1 = 0.05 mol;
11) n (CH3COOH) = n2 (KOH) = 0.05 mol;
12) m (CH3COOH) = n * M = 0.05 * 60 = 3 g.

Answer: The mass of C6H5OH is 9.4 g; CH3COOH – 3 g.