Question

9.3 \, \text{g of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is 100\% completed is } \\_\\_\\_\\_\\_ \times 10^{-1} \, \text{g.} \\\ \text{(Given molar mass in g mol}^{-1}\text{) N: 14, O: 16, C: 12, H: 1}

Answer 1

The reaction between aniline and acetic anhydride produces acetanilide. The balanced equation is:

$C_6H_5NH_2 + CH_3COOCOCH_3 \rightarrow C_6H_5NHCOCH_3 + CH_3COOH$

Given:
- Molar mass of aniline ($C_6H_7N$) = $93 \, \text{g/mol}$
- Molar mass of acetanilide ($C_8H_9NO$) = $135 \, \text{g/mol}$

Calculate moles of aniline:
$n_{\text{aniline}} = \frac{9.3}{93} = 0.1 \, \text{moles}$

Since the reaction is $1:1$, moles of acetanilide produced = moles of aniline = $0.1 \, \text{moles}$.

Mass of acetanilide produced:
$\text{Mass} = n \times \text{molar mass} = 0.1 \times 135 = 13.5 \, \text{g}$

Thus, $13.5 \, \text{g}$ or $135 \times 10^{-1} \, \text{g}$ of acetanilide is produced.
The Correct answer is: 135