In the given reaction, C6H6+Br2→C6H5Br+HBr 39gm of C6H6 reacts with 40gm of Br2. It was found that 19.625gm 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 A50% 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: =0.25mol=39.25gm %Yield=ActualYieldTheoriticalYield×100 ActualYield=19.625gm TheoriticalYield=39.25gm PercentageYield=19.62539.25×100=50%