Question
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) in a two-step process as shown here:
C6H5NH2 +HNO2 + HCl →C6H5N2+Cl− + 2H2O
C6H5N2+Cl− +KI→ C6H5I +N2 +KCl
In an actual preparation, 9.30 gof aniline was converted to 12.32 gof iodobenzene. The percentage yield of iodobenzene is:
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) in a balanced equation is shown in a two-step process:
C6H5NH2 +HNO2 + HCl →C6H5N2+Cl− + 2H2O
C6H5N2+Cl− +KI→ C6H5I +N2 +KCl
From the balanced equation, it is quite clear that 1 mole ofC6H5NH2, aniline generate 1 mole ofC6H5I, 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=6×12+5+127=204
1 mole of C6H5NH2(123g) = 1 mole of C6H5I(204g)
∴ 9.30 g aniline will give = 93204× 9.3 g= 20.4 g iodobenzene