1. Write the reaction

Benzoic acid reacts with sodium bicarbonate as:

$\mathrm{C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2}$

From the balanced equation,

• $1$ mole of benzoic acid produces $1$ mole of $\mathrm{CO_2}$.

2. Find moles of $\mathrm{CO_2}$ released

At STP, $1$ mole of any gas occupies $22.4\,\mathrm{L}$.

Given volume of $\mathrm{CO_2}$:

$11.2\,\mathrm{L}$

So, moles of $\mathrm{CO_2}$ are:

$n(\mathrm{CO_2}) = \frac{11.2}{22.4} = 0.5\,\mathrm{mol}$

3. Find moles of benzoic acid

Since the mole ratio is $1:1$,

$n(\text{benzoic acid}) = 0.5\,\mathrm{mol}$

4. Calculate molar mass of benzoic acid

Benzoic acid = $\mathrm{C_6H_5COOH = C_7H_6O_2}$

Molar mass:

$7\times 12 + 6\times 1 + 2\times 16 = 84 + 6 + 32 = 122\,\mathrm{g/mol}$

5. Calculate mass of benzoic acid

$m = n \times M = 0.5 \times 122 = 61\,\mathrm{g}$

So,

$X = 61\,\mathrm{g}$

6. Comparison with stored correct answer

Stored correct answer = $61$

Our derived answer is also $61$, so it agrees.