a
Reaction suggests that \(1\) mole of aniline give \(1\) mole of orange dye.

\(\text { so }(\mathrm{mol})_{\text {aniline }}=(\mathrm{mole})_{\text {orange dye }}\)

\(\frac{9.3 \mathrm{~g}}{93 \mathrm{~g} \mathrm{~mol}^{-1}}=\frac{\text { mass of orange dye }}{199 \mathrm{~g} \mathrm{~mol}^{-1}}\)

\(\text { mass of orange dye }=19.9 \mathrm{~g} \simeq 20 \mathrm{~g}\)