Question

In the given 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: 1.0 u, O: 16.0 u, N: 14.0 u)

Answer 1

1. moles = given mass / molar mass
2. Moles of C₆H₆ (benzene) =$\frac{givenmass}{molarmass}$ = $\frac{3.9}{78}$= 0.05 mol
3. Moles of C₆H₅NO₂ (nitrobenzene) = $\frac{givenmass}{molarmass}$=$\frac{4.92}{123}$ = 0.04 mol
4. theoretical yield is 0.05 mol
5. percentage yield = $\frac{experimentalyield}{theoreticalyield}\times 100$
6. percentage yield = $\frac{0.04}{0.05}\times 100$
7. percentage yield = 80

So, the correct answer is 80.