# Determine the mass of the precipitate and the volume of gas that forms when 0.2 mol

**Determine the mass of the precipitate and the volume of gas that forms when 0.2 mol of phenol reacts with an excess of bromine.**

We implement the solution:

1. By the condition of the problem, we write down the equation:

Y = 0.2 mol X g -? V -?

С6Н5ОН + 3Br2 = C6H3Br3 + 3HBr – substitutions, obtained tribromophenol in the sediment, hydrogen bromide;

2. Calculations:

M (C6H3Br3) = 314.7 g / mol;

Y (C6H5OH) = 0.2 mol;

Y (C6H3Br3) = 0.2 mol since the amount of substances according to the equation is 1 mol.

3. Find the mass of the product:

m (C6H3Br3) = Y * M = 0.2 * 314.7 = 62.94 g.

4. Proportion:

0.2 mol (C6H5OH) – X mol (HBr);

-1 mol -3 mol from here, X mol (HBr) = 0.2 * 3/1 = 0.6 mol.

5. Let’s calculate the volume of HBr:

V (HBr) = 0.6 * 22.4 = 13.44 liters.

Answer: tribromophenol was obtained with a mass of 62.94 g and hydrogen bromide with a volume of 13.44 liters.