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 \underline\hspace2cm %. (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)

Practical Organic Chemistry PYQ — JEE Main 2021 · 17 Mar · Shift 1 · Q17 with Solution

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Correct answer: 80

Write the nitration reaction

Benzene undergoes nitration as:
$\mathrm{C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O}$
From the equation, $1\text{ mol benzene } \rightarrow 1\text{ mol nitrobenzene}$
Calculate molar masses
- Benzene, $\mathrm{C_6H_6}$ :
  $6(12) + 6(1) = 72 + 6 = 78\,\text{g mol}^{-1}$
- Nitrobenzene, $\mathrm{C_6H_5NO_2}$ :
  $6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123\,\text{g mol}^{-1}$
Find moles of benzene used

Given mass of benzene $= 3.9\,\text{g}$
$n(\text{benzene}) = \frac{3.9}{78} = 0.05\,\text{mol}$
Find theoretical yield of nitrobenzene

Since the mole ratio is $1:1$ ,
$n(\text{nitrobenzene}) = 0.05\,\text{mol}$
Therefore theoretical mass of nitrobenzene is:
$m_{\text{theoretical}} = 0.05 \times 123 = 6.15\,\text{g}$
Calculate percentage yield

Actual mass of nitrobenzene obtained $= 4.92\,\text{g}$
$\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$ $\%\text{ yield} = \frac{4.92}{6.15} \times 100 = 80\%$
Final answer

Rounded to the nearest integer:
$\boxed{80}$
Comparison with stored answer

Stored correct answer = $80$

My derived answer matches the stored answer.