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.
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
Match the Compounds (List - I) with the appropriate Catalyst/Reagents (List - II) for their reduction into corresponding amines. 
Consider the following amino acid: 
Which of the following options contain the correct structure of (A) and (B)?