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Balancing Redox Equations

In the above ...

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 :1.0 u, O : 16.0 u, N : 14. 0 u)

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Solution

$Moles of C_{6} H_{6} = \frac{3.9}{78} = 0.05 m o l$
$Moles of C_{6} H_{5} N O_{2} = \frac{4.92}{123} = 0.04 m o l$
By conserving moles of carbon, mole of $C_{6} H_{5} N O_{2}$ formed theoreticallyshould be 0.05 mol.
$% Y i e l d = \frac{m o l e s f o r m e d a c t u a l l y}{m o l e s f o r m e d t h e o r e t i c a l l y} \times 100$

$% Y i e l d = \frac{0.04}{0.05} \times 100 = 80 %$

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