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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. 0.08B. 0.5C. 0.75D. 0.8 |
Answer» 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%` |
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