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 :-

We need to find the percentage yield of acetanilide preparation from aniline.

The reaction is $$C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$$.

We are given that the mass of aniline is 9.3 g and its molar mass (C₆H₅NH₂) is 93 g/mol, which means the moles of aniline are 9.3/93 = 0.1 mol.

Since the stoichiometry is 1:1, the theoretical moles of acetanilide formed are also 0.1 mol. Using its molar mass (C₆H₅NHCOCH₃) of 135 g/mol, the theoretical mass of acetanilide is 0.1 × 135 = 13.5 g.

The actual yield obtained in the experiment is 11 g.

Calculating the percentage yield gives $$\% \text{ yield} = \frac{11}{13.5} \times 100 = 81.48\% \approx 81.5\%$$.

Therefore, the percentage yield is Option 2: 81.5%.