Iodobenzene is prepared from aniline `(C_(6)H_(5)NH_(2))` in a two step process as shown here: `C_(6)H_(5)NH_(2)+HNO_(2)+HCl to C_(6)H_(5)N_(2)^(+)Cl+

Question

Iodobenzene is prepared from aniline `(C_(6)H_(5)NH_(2))` in a two step process as shown here: `C_(6)H_(5)NH_(2)+HNO_(2)+HCl to C_(6)H_(5)N_(2)^(+)Cl+

Iodobenzene is prepared from aniline `(C_(6)H_(5)NH_(2))` in a two step process as shown here:
`C_(6)H_(5)NH_(2)+HNO_(2)+HCl to C_(6)H_(5)N_(2)^(+)Cl+2H_(2)O`
`C_(6)H_(5)N_(2)^(+)Cl^(-)+KI to C_(6)H_(5)I+N_(2)+KCl`
In an actual preparation, 9.30g of aniline was converted to 12.32g of iodobenzene. The percentage yield of iodobenzene is:
A. 0.08
B. 0.5
C. 0.75
D. 0.8

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - D
`1"mole of "C_(6)H_(5)NH_(2)(123g)-=1"mole of "C_(6)H_(5)I(204g)`
`therefore "9.3g of aniline will give"=((204)/(123)xx9.3)"g iodobenzene"`
=15.24g iodobenzene
`"% yield"=("Actual amount of product")/("Calculated amount of product")xx100`
`=(12.32)/(15.424)xx100=80%`

6 views Answered 2023-02-05 15:29