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$5 \mathrm{~g}$ of benzene on nitration gave $6.6 \mathrm{~g}$ of nitrobenzene. The theoretical yield of the nitrobenzene will be
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$8.09 \mathrm{~g}$
$8.09 \mathrm{~g}$
$\mathrm{C}_6 \mathrm{H}_6+\mathrm{HNO}_3 \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2+\mathrm{H}_2 \mathrm{O}$ $78 \mathrm{~g} \quad 123 \mathrm{~g}$ Now since $78 \mathrm{~g}$ of benzene on nitration give $=123 \mathrm{~g}$ nitrobenzene hence $5 \mathrm{~g}$ of benzene on nitration give $=\frac{123}{78} \times 5=7.88 \mathrm{~g}$
nearest answer is (c) i.e. theoritical yield $=7.88 \mathrm{~g}$
nearest answer is (c) i.e. theoritical yield $=7.88 \mathrm{~g}$
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