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Question: Iodobenzene is prepared from aniline \[\left( {{C_6}{H_5}N{H_2}} \right)\] in a two-step process as ...

Iodobenzene is prepared from aniline (C6H5NH2)\left( {{C_6}{H_5}N{H_2}} \right) in a two-step process as shown here:
C6H5NH2 +HNO2 + HCl C6H5N2+Cl + 2H2O{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
C6H5N2+Cl +KI C6H5I +N2 +KCl{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, 9.30 g{\text{9}}{\text{.30 g}}of aniline was converted to 12.32 g12.32{\text{ }}gof iodobenzene. The percentage yield of iodobenzene is:

A.8 % B.50 % C.25 % D.60 %  A.8{\text{ }}\% \\\ B.50{\text{ }}\% \\\ C.25{\text{ }}\% \\\ D.{\text{}}60{\text{ }}\% \\\
Explanation

Solution

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 (C6H5NH2)\left( {{C_6}{H_5}N{H_2}} \right) in a balanced equation is shown in a two-step process:
C6H5NH2 +HNO2 + HCl C6H5N2+Cl + 2H2O{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
C6H5N2+Cl +KI C6H5I +N2 +KCl{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 ofC6H5NH2  {C_6}{H_5}_N{H_2}\;, aniline generate 1 mole ofC6H5I{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)$$: 6C+5H+ 127=:{\text{ }}6C + 5H + {\text{ }}127 = 6×12+5+127=2046 \times 12 + 5 + 127 = 204
1 mole of C6H5NH2(123g)  {C_6}{H_5}_N{H_2}\left( {123g} \right)\; = 1 mole of C6H5I(204g){C_6}{H_5}I\left( {204g} \right)
9.30 g9.30{\text{ }}g aniline will give = 20493× 9.3 g  = 20.4 g\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.