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Q.

$C_{6} H_{6} + {Br}_{2} ⟶ C_{6} H_{5} Br + HBr$
In an experiment, to prepare bromobenzene according to the equation, a student reacted 20.0 g of $C_{6} H_{6}$ with 0.310 mol of bromine. If 28.0 g of $C_{6} H_{5} Br$ was obtained, what was the percentage yield?

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a

31.5

b

57.6

c

40.3

d

69.7

answer is D.

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Detailed Solution

In the reaction we have,

$C_{6} H_{6} + {Br}_{2} ⟶ C_{6} H_{5} Br + HBr$

$n = 0.256 0.310$

$(LR) = \frac{0.178}{0.256} \times 100 = 69.7 % yield$

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