Question #3fc19

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Answer 1 · 2017-06-08T15:40:43

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%#