Question:

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

Updated On: Nov 3, 2025
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Correct Answer: 135

Approach Solution - 1

To find the mass of acetanilide formed, we first calculate the moles of aniline. The molecular formula for aniline is C6H7N. Using the given atomic masses:

C: 6 × 12= 72
H: 7 × 1= 7
N: 1 × 14= 14

The molar mass of aniline is 72 + 7 + 14 = 93 g/mol. Given the mass of aniline is 9.3 g, the moles of aniline is calculated as:

moles of aniline = 9.3/93 = 0.1 mol 

The reaction between aniline and acetic anhydride forms acetanilide (C8H9NO). The molar mass of acetanilide is:

C: 8 × 12= 96
H: 9 × 1= 9
N: 1 × 14= 14
O: 1 × 16= 16

Molar mass of acetanilide = 96 + 9 + 14 + 16 = 135 g/mol

Assuming the reaction is 100% efficient, the moles of acetanilide formed equals the moles of aniline reacted, thus: 0.1 mol of acetanilide is formed.

The mass of acetanilide produced is:

mass = 0.1 × 135 = 13.5 g

Given the format of the answer as __ × 10-1 g, we express 13.5 g as:

13.5 = 135 × 10-1 g

The result of 135 falls within the specified range of 135—135, confirming accuracy.

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Approach Solution -2

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

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