JEE Chemistry 2024 Question with Solution

Given: Mass of $\text{CO}_2$ produced is $22 \, \text{g}$.

Find: Mass of $\text{CH}_4$ required for complete combustion.

The balanced chemical equation is

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$

Molar mass of $\text{CO}_2$ is

$$12 + 2(16) = 44 \, \text{g/mol}$$

Moles of $\text{CO}_2$ produced are

$$\frac{22}{44} = 0.5 \, \text{mol}$$

From the balanced equation, $1$ mole of $\text{CH}_4$ produces $1$ mole of $\text{CO}_2$. Therefore, required moles of $\text{CH}_4$ are

$$0.5 \, \text{mol}$$

Molar mass of $\text{CH}_4$ is

$$12 + 4(1) = 16 \, \text{g/mol}$$

Hence, mass of $\text{CH}_4$ required is

$$0.5 \times 16 = 8 \, \text{g}$$

Therefore, the required mass of methane is $8 \, \text{g}$. The answer key shows $44 \, \text{g}$, but the extracted solution clearly gives $8 \, \text{g}$, so the answer is $8$.

Given: $\text{CH}_4$ and $\text{CO}_2$ are in a $1:1$ mole ratio in complete combustion.

Find: Mass of $\text{CH}_4$ needed for $22 \, \text{g}$ of $\text{CO}_2$.

Since $44 \, \text{g}$ of $\text{CO}_2$ corresponds to $1$ mole, $22 \, \text{g}$ corresponds to $0.5$ mole. Therefore, methane required is also $0.5$ mole.

Now convert moles of $\text{CH}_4$ to mass:

$$0.5 \times 16 = 8 \, \text{g}$$

Therefore, the correct numerical answer is $8$.