Mole ConceptMedium
Question
Iodobenzene is prepared from aniline (C6H5NH2) in a two-step process as shown here.
C6H5NH2+HNO2+HCl?C6H5N2Cl+2H2O
C6H5N2Cl+KI? C6H5I+N2+KCl
In an actual preparation, 9.30 g of aniline was converted to 16.32 g of iodobenzene. The percentage yield of iodobenzene is (I = 127)
Options
A.8%
B.50%
C.75%
D.80%
Solution
$C_{6}H_{5}NH_{2} + HNO_{2} + HCl \rightarrow C_{6}H_{5}N_{2}^{+}Cl^{-} + 2H_{2}O$
$C_{6}H_{5}N_{2}^{+}Cl^{-} + KI \rightarrow C_{6}H_{5}I + N_{2} + KCl$
Theoretical yield of C6H5I = $\frac{204}{93} \times 9.3 = 20.4\text{ g}$
Hence, the percentage yield = $\frac{16.32}{20.4} \times 100 = 80\%$
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