$$ X\,g $$ of benzoic acid on reaction with aq. $$NaHCO_3$$ released $$CO_2$$ that occupied $$11.2\,L$$  volume at STP.  $$X$$  is  $$\underline{\hspace{2cm}}\,g. $$

We need to find the mass of benzoic acid that produces $$CO_2$$ occupying 11.2 L at STP.

Benzoic acid reacts with aqueous sodium bicarbonate: $$C_6H_5COOH + NaHCO_3 \rightarrow C_6H_5COONa + H_2O + CO_2 \uparrow$$ The stoichiometry shows that 1 mole of benzoic acid produces 1 mole of $$CO_2$$.

At STP (Standard Temperature and Pressure: 0 degC, 1 atm), 1 mole of any ideal gas occupies 22.4 L. $$\text{Moles of } CO_2 = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ mol}$$

From the 1:1 stoichiometry: $$\text{Moles of benzoic acid} = \text{Moles of } CO_2 = 0.5 \text{ mol}$$

The molar mass of benzoic acid ($$C_6H_5COOH = C_7H_6O_2$$) is $$(7 \times 12) + (6 \times 1) + (2 \times 16) = 84 + 6 + 32 = 122 \text{ g/mol}$$

$$X = 0.5 \times 122 = 61 \text{ g}$$

The answer is 61 g.