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93 g of aniline produces 330 g of 2,4,6-tribromoaniline.
Theoretical yield for 9.3 g of aniline:
\[ \frac{330}{93} \times 9.3 = 33 \, \text{g} \]
Percentage yield:
\[ \text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 \]
\[ \text{Percentage yield} = \frac{26.4}{33} \times 100 = 80\% \]
During "S" estimation, 160 mg of an organic compound gives 466 mg of barium sulphate. The percentage of Sulphur in the given compound is %.
(Given molar mass in g mol\(^{-1}\) of Ba: 137, S: 32, O: 16)