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\(9.3 \mathrm{~g}\) of pure aniline is treated with bromine water at room temperature to give a white precipitate of the product ' \(\mathrm{P}\) '. The mass of product ' \(\mathrm{P}\) ' obtaind is \(26.4 \mathrm{~g}\). The percentage yield is _______ \(\%\).
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80
$93 \mathrm{~g}$ of aniline produces $330 \mathrm{~g}$ of $2,4,6-$ tribromoaniline. Hence $9.3 \mathrm{~g}$ of aniline should produce $33 \mathrm{~g}$ of 2, 4, 6-tribromoaniline. Hence percentage yield $\frac{26.4 \times 100}{33}=80 \%$
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