JEE Chemistry 2023 Question with Solution

Given: Mass of cyclohexanecarbaldehyde = $131.8 \, \text{kg}$

Find: Mass of $\mathrm{NH_3}$ produced.

The chemical equation for the reaction is

$$\text{Cyclohexanecarbaldehyde} + 2[\mathrm{Ag(NH_3)_2}]\mathrm{OH} \rightarrow \text{Cyclohexanecarboxamide} + 3\mathrm{NH_3} + 2\mathrm{Ag} + \mathrm{H_2O}$$

The molecular weight of cyclohexanecarbaldehyde is taken as

$$12 \times 8 + 14 + 16 = 112 \, \text{g/mol}$$

The number of moles of cyclohexanecarbaldehyde is

$$\frac{131.8 \times 1000}{112} = 1176.8 \, \text{mol}$$

From the stoichiometry, $1$ mole of cyclohexanecarbaldehyde produces $3$ moles of $\mathrm{NH_3}$. So,

$$\text{Moles of } \mathrm{NH_3} = 3 \times 1176.8 = 3530.4 \, \text{mol}$$

The molar mass of $\mathrm{NH_3}$ is

$$17 \, \text{g/mol}$$

Therefore, the mass of $\mathrm{NH_3}$ produced is

$$3530.4 \times 17 = 60023.2 \, \text{g} = 60 \, \text{kg}$$

Therefore, the required mass of $\mathrm{NH_3}$ is $60 \, \text{kg}$, so the numerical answer is 60.

Given: Cyclohexanecarbaldehyde produces $3$ moles of $\mathrm{NH_3}$ per mole.

Find: Mass of $\mathrm{NH_3}$ directly from mass ratio.

Use the mole ratio and molar masses together:

$$112 \, \text{g of cyclohexanecarbaldehyde} \rightarrow 3 \times 17 = 51 \, \text{g of } \mathrm{NH_3}$$

So,

$$131.8 \, \text{kg} \rightarrow 131.8 \times \frac{51}{112} \, \text{kg}$$ $$= 60.0232 \, \text{kg}$$

Nearest integer = 60.

This works because stoichiometry lets us convert directly from reactant mass to product mass using the fixed mole ratio.