JEE Main 2024 — Some Basic Concepts Of Chemistry Question with Solution

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 $$95.04\times 10^{-1}\mathrm{g}$$.