b
\(93 \mathrm{~g}\) 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 \(\frac{26.4 \times 100}{33}=80 \%\)
