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Question

In the given reaction, C6H6 + Br2 C6H5Br + HBr
39 gm of C6H6 reacts with 40 gm of Br2. It was found that 19.625 gm of C6H5Br was obtained. Calculate the percentage yield of the reaction.

A
50 %
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B
85 %
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C
42.5 %
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D
33.3 %
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Solution

The correct option is A 50 %
In this reaction, C6H6 + Br2 C6H5Br + HBr
Moles of C6H6, Br2, C6H5Br :
nC6H6=3978=0.5nBr2=40160=0.25nC6H5Br=19.625157=0.125

We need to find the limiting reagent here, which is the one whose ratio of actual moles and stoichiometric coefficients is the smallest.
Br2 is the limiting reagent for this reaction.
From reaction stoichiometry:
Expected C6H5Br formed:
=0.25 mol=39.25 gm
% Yield=Actual YieldTheoritical Yield× 100
Actual Yield=19.625 gm
Theoritical Yield=39.25 gm
Percentage Yield=19.62539.25×100=50%

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