Question

Iodobenzene is prepared from aniline \[\left( {{C_6}{H_5}N{H_2}} \right)\] in a two-step process as shown here:
\[{C_6}{H_5}N{H_2}{\text{ }} + HN{O_2}{\text{ }} + {\text{ }}HCl{\text{ }} \to {C_6}{H_5}{N_2}^ + C{l^ - }{\text{ }} + {\text{ }}2{H_2}O\]
\[{C_6}{H_5}{N_2}^ + C{l^ - }{\text{ }} + KI \to {\text{ }}{C_6}{H_5}I{\text{ }} + {N_2}{\text{ }} + KCl\]
In an actual preparation, \[{\text{9}}{\text{.30 g}}\]of aniline was converted to \[12.32{\text{ }}g\]of iodobenzene. The percentage yield of iodobenzene is:
\[
  A.8{\text{ }}\% \\
  B.50{\text{ }}\% \\
  C.25{\text{ }}\% \\
  D.{\text{}}60{\text{ }}\% \\
 \]

Answer 1

Hint: In order to solve this question we need to understand the concept of ‘mole’. A mole of a substance or a mole of particles is defined as exactly particles, which may be atoms, molecules, ions, or electrons. Now, it is interesting to note that there is a relationship between mole and molar mass. The molar mass is the mass in grams of 1 mole of substance.

Complete step by step solution:
The reaction of preparation of iodobenzene from aniline \[\left( {{C_6}{H_5}N{H_2}} \right)\] in a balanced equation is shown in a two-step process:
\[{C_6}{H_5}N{H_2}{\text{ }} + HN{O_2}{\text{ }} + {\text{ }}HCl{\text{ }} \to {C_6}{H_5}{N_2}^ + C{l^ - }{\text{ }} + {\text{ }}2{H_2}O\]
\[{C_6}{H_5}{N_2}^ + C{l^ - }{\text{ }} + KI \to {\text{ }}{C_6}{H_5}I{\text{ }} + {N_2}{\text{ }} + KCl\]
From the balanced equation, it is quite clear that 1 mole of\[{C_6}{H_5}_N{H_2}\;\], aniline generate 1 mole of\[{C_6}{H_5}I\], iodobenzene.
Now, we need to find the mass of these molar mass of these compounds:
Molar mass of aniline \[\left( {{C_6}{H_5}N{H_2}} \right)\]\[:{\text{ }}6C + {\text{ }}5 + 1N + 2H = \]$6 \times 12 + 5 + 14 + 2 = 93$
Molar mass of iodobenzene \[({C_6}{H_5}I)\]\[:{\text{ }}6C + 5H + {\text{ }}127 = \]$6 \times 12 + 5 + 127 = 204$
1 mole of \[{C_6}{H_5}_N{H_2}\left( {123g} \right)\;\] = 1 mole of \[{C_6}{H_5}I\left( {204g} \right)\]
∴ \[9.30{\text{ }}g\] aniline will give = \[\dfrac{{204}}{{93}} \times {\text{ }}9.3{\text{ }}g\; = {\text{ }}20.4{\text{ }}g\] iodobenzene
\[
  \% {\text{ }}Yield\; = \dfrac{{Actual{\text{ }}amount{\text{ }}of{\text{ }}product}}{{Theoretical{\text{ }}amount{\text{ }}of{\text{ }}product}} \times 100 \\
  \% {\text{ }}Yield\; = \dfrac{{12.32}}{{20.4}} \times 100 = 60\% \\
 \] ​
Therefore, we will get 60% yield of iodobenzene from aniline.

Hence, answer is D. 60%

Note: We can also know that the WHO (World Health Organization) suggest six litres of water per person daily to meet the requirements of most people under most conditions; and around 15 litres per person daily to cover basic hygiene and food hygiene needs.