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

Correct Answer: 135

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