In the above reaction, $3.9 \mathrm{~g}$ of benzene on nitration gives $4.92 \mathrm{~g}$ of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _________ %. (Round off to the Nearest Integer).
Given atomic mass : $\mathrm{C}: 12.0 \mathrm{u}, \mathrm{H}: 1.0 \mathrm{uO}: 16.0 \mathrm{u}, \mathrm{N}: 14.0 \mathrm{u}$
(80)
$\begin{array}{ll}1 \text { mole } & 1 \text { mole } \\ 78 \mathrm{gm} & 123 \mathrm{gm}\end{array}$
$3.9 \mathrm{gm} \quad \frac{123}{78} \times 3.9=6.15 \mathrm{gm}$
But actual amount of nitrobenzene formed is $4.92 \mathrm{gm}$ and hence.
Percentage yield $=\frac{4.92}{6.15} \times 100=80 \%$
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