# 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.

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