Aniline weighing 27.9 g reacts with an excess of bromine, determine the mass of the resulting 2,4,6-tribromaniline.

Let’s write the reaction equation:

C6H5NH2 + 3Br2 = C6H2Br3NH2 ↓ + 3HBr

Let’s find the amount of aniline substance:

v (C6H5NH2) = m (C6H5NH2) / M (C6H5NH2) = 27.9 / 93 = 0.3 (mol).

According to the reaction equation, from 1 mol of C6H5NH2, 1 mol of C6H2Br3NH2 is formed, therefore:

v (C6H2Br3NH2) = v (C6H5NH2) = 0.3 (mol).

Thus, the mass of the resulting precipitate of 2,4,6-tribromaniline:

m (C6H2Br3NH2) = v (C6H2Br3NH2) * M (C6H2Br3NH2) = 0.3 * 330 = 99 (g).

Answer: 99 g.