
Reaction suggests that 1 mole of aniline give 1 mole of orange dye. $\begin{aligned}
& \text { so }(\mathrm{mol}){\text {aniline }}=(\mathrm{mole}){\text {orange dye }} \
& \frac{9.3 \mathrm{g}}{93 \mathrm{g} \mathrm{mol}^{-1}}=\frac{\text { mass of orange dye }}{199 \mathrm{g} \mathrm{mol}^{-1}}
\end{aligned}massoforangedye=19.9 \mathrm{g} \simeq 20 \mathrm{~g}$