Question

Iodobenzene $\left(C_6{H}_{5}I\right)$ is prepared from aniline $\left(C_6{H}_{5}N{H}_2\right)$ in a two step process as shown below:

$C_6{H}_{5}N{H}_2+HN{O}_2+HCl\to C_6{H}_5{N}_2{}^{+}C{l}^{-}+2H_{2}O$

$C_6{H}_5{N}_2{}^{+}C{l}^{-}+KI\to C_6{H}_{5}I+N_2+KCl$

In an actual preparation, $9.30$ g of aniline was converted to $16.32$ g of iodobenzene. The percentage yield of iodobenzene is:

• A. $8\%$
• B. $50\%$
• C. $75\%$
• D. $80\%$

Answer 1

The correct option is D $80\%$

$9.30gm$ Aniline $\to 16.32g$ of Iodobenzene

$CC{H}_{5}N{H}_2+HN{O}_2+HCl\to C_6{H}_5{N}_2+Cl+2H_{2}O$

$93gm190.5gm$

$9.30\to xgm$

$x=\frac{140.5\times 9.30}{93}$

$=14.05$

$C_6{H}_5{N}_2^{+}+C{l}^{-}+KI\to C_6{H}_{5}I+N_2+KCl$

$140.5gm\to 207$

$14.05\to ygm$

$y=\frac{207\times 14.05}{140.5}$

$y=20.7$ ..............Theoretical Yield

Practical Yield $=\frac{ActualYield}{Theoretical}\times 100$

$=\frac{16.32}{20.7}\times 100$

$=78.84\approx 79\%$

Hence, the correct option is $D$