Question

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 : 12.0 u, H : 1.0u, O : 16.0 u, N : 14.0 u)

Answer 1

But actual amount of nitrobenzene formed is 4.92 gm and hence.

Percentage yield = \(\frac{4.92}{6.15}\times100\) = 80%