$$9.3 \text{ g}$$ of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product '$$P$$'. The mass of product '$$P$$' obtained is $$26.4 \text{ g}$$. The percentage yield is ______ %.

Aniline reacts with bromine water to give 2,4,6-tribromoaniline (white precipitate).

Moles of aniline.

Molar mass of aniline = 93 g/mol. Moles = $$\frac{9.3}{93} = 0.1$$ mol.

Theoretical yield of 2,4,6-tribromoaniline.

Molar mass of $$C_6H_2Br_3NH_2$$ = 72 + 4 + 14 + 240 = 330 g/mol.

Theoretical yield = $$0.1 \times 330 = 33$$ g.

Percentage yield.

The answer is 80.