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JEE Chemistry 2026 Question with Solution

The mass of benzanilide obtained from the benzoylation reaction of 5.8 g5.8 \, g of aniline, if yield of product is 82%82\%, is _____ gg (nearest integer).

(Given molar mass in g  mol−1g \; mol^{-1}: H : 11, C : 1212, N : 1414, O : 1616)

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: Mass of aniline =5.8 g= 5.8 \, g, percentage yield =82%= 82\%.

Find: Mass of benzanilide obtained, nearest integer.

Step 1: Write the reaction stoichiometry.

Aniline reacts with benzoyl chloride to form benzanilide in a 1:11:1 molar ratio.

Step 2: Calculate molar mass of aniline.

Aniline: C6H5NH2=C6H7N\mathrm{C_6H_5NH_2} = \mathrm{C_6H_7N}

Molar mass of aniline=(6×12)+(7×1)+14=93 g mol−1\text{Molar mass of aniline} = (6 \times 12) + (7 \times 1) + 14 = 93 \, \text{g mol}^{-1}

Step 3: Calculate number of moles of aniline.

Moles of aniline=5.893≈0.062 mol\text{Moles of aniline} = \frac{5.8}{93} \approx 0.062 \, \text{mol}

Step 4: Calculate molar mass of benzanilide.

Benzanilide: C13H11NO\mathrm{C_{13}H_{11}NO}

Molar mass=(13×12)+(11×1)+14+16=197 g mol−1\text{Molar mass} = (13 \times 12) + (11 \times 1) + 14 + 16 = 197 \, \text{g mol}^{-1}

Step 5: Calculate theoretical mass of benzanilide.

Theoretical mass=0.062×197≈12.2 g\text{Theoretical mass} = 0.062 \times 197 \approx 12.2 \, \text{g}

Step 6: Apply percentage yield.

Actual mass=82100×12.2≈10.0 g\text{Actual mass} = \frac{82}{100} \times 12.2 \approx 10.0 \, \text{g}

Therefore, the mass of benzanilide obtained is 10 g10 \, g, so the required numerical answer is 10.

Yield-Based Stoichiometric Approach

Given: The reaction produces benzanilide from aniline with percentage yield 82%82\%.

Find: The actual mass of product formed.

First use the given atomic masses to compute the molar masses of reactant and product. Then convert the given mass of aniline into moles, use the 1:11:1 stoichiometric ratio, and finally apply the percentage yield to the theoretical product mass.

Aniline=C6H7NM=6(12)+7(1)+14=93 g mol−1\begin{aligned} \text{Aniline} &= \mathrm{C_6H_7N} \\ M &= 6(12) + 7(1) + 14 = 93 \, \text{g mol}^{-1} \end{aligned}
n(aniline)=5.893≈0.062 mol\begin{aligned} n(\text{aniline}) &= \frac{5.8}{93} \\ &\approx 0.062 \, \text{mol} \end{aligned}

Because the molar ratio is 1:11:1, moles of benzanilide formed theoretically are also 0.062 mol0.062 \, \text{mol}.

Benzanilide=C13H11NOM=13(12)+11(1)+14+16=197 g mol−1\begin{aligned} \text{Benzanilide} &= \mathrm{C_{13}H_{11}NO} \\ M &= 13(12) + 11(1) + 14 + 16 = 197 \, \text{g mol}^{-1} \end{aligned}
Theoretical mass of benzanilide=0.062×197≈12.2 g\begin{aligned} \text{Theoretical mass of benzanilide} &= 0.062 \times 197 \\ &\approx 12.2 \, \text{g} \end{aligned}
Actual mass=82100×12.2≈10.0 g\begin{aligned} \text{Actual mass} &= \frac{82}{100} \times 12.2 \\ &\approx 10.0 \, \text{g} \end{aligned}

Hence, the nearest integer value is 10.

Common mistakes

  • Using the given 5.8 g5.8 \, g directly as product mass is incorrect because percentage yield must be applied to the theoretical yield of benzanilide, not to the reactant mass. First convert aniline to moles, then compute the product mass.

  • Calculating the molar mass of aniline incorrectly by missing one hydrogen is a common error. Aniline is C6H7N\mathrm{C_6H_7N}, so its molar mass is 93 g mol−193 \, \text{g mol}^{-1}, not a nearby incorrect value.

  • Forgetting the 1:11:1 stoichiometric ratio between aniline and benzanilide leads to an incorrect mole conversion. Use the reaction stoichiometry before converting moles of reactant into mass of product.

  • Applying 82%82\% as 8282 instead of 82100\frac{82}{100} gives a result larger than the theoretical yield, which is impossible. Percentage yield must always reduce the theoretical mass when it is below 100%100\%.

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