QJEEM24Ch6bcd: JEE Main 2024 Chemistry - Class 11 JEEM24Ch6bcd Solution: From of aniline, the maximum amount of acetanilide that can be…

To determine the maximum amount of acetanilide that can be prepared from 6.55 g of aniline, we need to use stoichiometry. Let's go through the process step by step.

1. Molecular weights calculation:

The molecular weight of aniline (C₆H₅NH₂) is calculated as follows:

$\text{Molecular weight of aniline} = 6 \times 12 + 5 \times 1 + 14 + 2 \times 1 = 93 \text{ g/mol}$

The molecular weight of acetanilide (C₈H₉NO) is calculated as follows:

$\text{Molecular weight of acetanilide} = 8 \times 12 + 9 \times 1 + 14 + 16 = 135 \text{ g/mol}$

2. Mole calculation:

Moles of aniline:

$\text{Moles of aniline} = \frac{\text{Mass of aniline}}{\text{Molecular weight of aniline}} = \frac{6.55 \text{ g}}{93 \text{ g/mol}} = 0.0704 \text{ mol}$

3. Stoichiometry of the reaction:

The reaction between aniline and acetic anhydride to form acetanilide follows a 1:1 mole ratio.

4. Mass calculation:

Theoretical mass of acetanilide formed:

$\text{Mass of acetanilide} = \text{Moles of aniline} \times \text{Molecular weight of acetanilide} = 0.0704 \text{ mol} \times 135 \text{ g/mol} = 9.504 \text{ g}$

5. Convert to the desired unit:

Given the unit required is $\times 10^{-1} \mathrm{~g}$ , we express 9.504 g as:

$9.504 \text{ g} = 95.04 \times 10^{-1} \text{ g}$

Therefore, the maximum amount of acetanilide that can be prepared from 6.55 g of aniline is $95.04 \times 10^{-1} \mathrm{~g}$ .