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)

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