Iodobenzene `(C_(6)H_(5)l)` is prepared from aniline `(C_(6)H_(5)NH_(2))` in a two step process as shown below `C_(6)H_(5)NH_(2) + HNO_(2) + HCl rarr C_(6)H_(5)N_(2) .^(+)Cl^(-) + 2H_(2)O " "C_(6)H_(5)N_(2) .^(+)Cl^(-) + KI rarr C_(6)H_(5)I + N_(2) + KCl`
In an actual preparation `9.30 g` of aniline was coverted to `16.32 g` of iodobenzene. The percentage yield of iodobenzene is :

Options

A. `8%`
B. `50%`
C. `75%`
D. `80%`

Answer

D

Explanation

`C_(6)H_(5)NH_(2)+HNO_(2)+HCIrarr C_(6)H_(5)N_(2).^(+)CI^(-)+2H_(2)O`
`C_(6)H_(5)N_(2).^(.+)CI^(-)+KIrarr C_(6)H_(5)I+N_(2)+KCl`
`n_(P)=n_(r)xxR_(1)xxR_(2)`
moles of `C_(6)H_(5)I`=mole of `C_(6)H_(5)NH_(2)xxR_(1)xxR_(2)`
`(wt)/(204)=(9.3)/(93)xx1xx1`
`wt=20.4 g`
`%` yield of `C_(6)H_(5)I`
`=(16.32)/(20.4)xx100=80%`