Question

Question #3fc19

Answer 1

Here is a conversion

`C_6H_5NH_2->C_6H_5I`

Molar masses are

`C_6H_5NH_2->6xx12+7xx1+1xx14=93g"/"mol`

`C_6H_5I->6xx12+5xx1+1xx127=204g"/"mol`

So theoretically `9.3`gm `C_6H_5NH_2` will produce `20.4g` `C_6H_5I` but `16.32g` is produced practically.

So

The percentage Yield

`="mass of product practically"/"mass of the product theoretically"xx100%`

`=16.32/20.4xx100%=80%`