In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is _______ % (Round off to the nearest integer).
(Given atomic mass : C: 12.0u, H:1.0u, O:16.0u, N:14.0u)

<p><img ></p><p>Moles of benzene = 3.9 g / 78 g/mol . This would produce (3.9 / 78) moles of nitrobenzene in 100% conversion.<br>Produced moles of nitrobenzene = 4.92 g / 123 g/mol .<br>% yield = [ (4.92 / 123) / (3.9 / 78) ] × 100 = [ (4.92 × 100 × 78) / (123 × 3.9) ] = 80.0%<br>Ans = 80</p>

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