The interaction of 18.8 g of phenol with an excess of bromine water formed 18 g of a precipitate

The interaction of 18.8 g of phenol with an excess of bromine water formed 18 g of a precipitate. Determine the product yield from the theoretically possible.

С6Н5ОН + 3Br2 = C6H2Br3OH + 3HBr – substitutions, tribromophenol precipitate was formed;
Calculations:
M (C6H5OH) = 94 g / mol;

M (C6H2Br3OH) = 330.7 g / mol.

Determine the amount of the original substance:
Y (C6H5OH) = m / M = 18.8 / 94 = 0.2 mol;

Y (C6H2Br3OH) = 0.2 mol since the amount of substances is 1 mol.

We find the mass of the sediment, the product yield:
m (C6H2Br3OH) = Y * M = 0.2 * 330.7 = 66.14 g (theoretical weight);

W = m (practical) / m (theoretical) * 100;

W = 18 / 66.14 * 100 = 27, 21%

Answer: the yield of tribromophenol is 27.21%