Question

A student has planned to prepare acetanilide from aniline using acetic anhydride. The student has started from 9.3 g of aniline. However, the student has managed to obtain 11 g of dry acetanilide. The % yield of this reaction is

• A. 97.5%
• B. 81.5%
• C. 59.5%
• D. 72.5%

Hint:

Always calculate percentage yield using actual yield over theoretical yield.

Answer 1

The Correct Option is B

The question involves calculating the percentage yield of acetanilide obtained from aniline using acetic anhydride. Let us solve this problem step-by-step.

1. Determine the Molar Masses:
   • Molar mass of aniline (C₆H₅NH₂): \(6 \times 12\,(\text{Carbon}) + 5 \times 1\,(\text{Hydrogen}) + 14\,(\text{Nitrogen}) + 1 \times 1\,(\text{Hydrogen}) = 93\,\text{g/mol}\)
   • Molar mass of acetanilide (C₈H₉NO): \(8 \times 12\,(\text{Carbon}) + 9 \times 1\,(\text{Hydrogen}) + 1 \times 14\,(\text{Nitrogen}) + 1 \times 16\,(\text{Oxygen}) = 135\,\text{g/mol}\)
2. Calculate the Theoretical Yield:
   • The balanced equation for the reaction is: \[ \text{C}_6\text{H}_5\text{NH}_2 + \text{(CH}_3\text{CO)}_2\text{O} \rightarrow \text{C}_8\text{H}_9\text{NO} + \text{CH}_3\text{COOH} \]
   • Given 9.3 g of aniline corresponds to \(\frac{9.3}{93} = 0.1\,\text{mol}\).
   • The reaction produces acetanilide with a 1:1 molar ratio, thus theoretically producing 0.1 mol of acetanilide.
   • Theoretical yield of acetanilide = \(0.1\, \text{mol} \times 135\,\text{g/mol} = 13.5\,\text{g}\).
3. Calculate the Percentage Yield:
   • Actual yield of acetanilide = 11 g.
   • Percentage yield = \(\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%\)
   • Substitute the values: \(\frac{11}{13.5} \times 100\% \approx 81.5\%\).

Therefore, the percentage yield of acetanilide in this reaction is 81.5%. This matches the given correct answer.