Xg of benzoic acid on reaction with aq. NaHCO3 released CO2 that occupied 11.2 L volume at STP. X is — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Write the balanced chemical equation** C₆H₅COOH + NaHCO₃ → C₆H₅COONa + H₂O + CO₂ Benzoic acid reacts with sodium bicarbonate in a 1:1 molar ratio to produce CO₂. **Step 2: Calculate moles of CO₂ produced** At STP, 1 mole of gas occupies 22.4 L Moles of CO₂ = Volume / Molar volume = 11.2 L / 22.4 L/mol = 0.5 mol **Step 3: Determine moles of benzoic acid** From the balanced equation, the mole ratio is: C₆H₅COOH : CO₂ = 1 : 1 Therefore, moles of benzoic acid = 0.5 mol **Step 4: Calculate molar mass of benzoic acid** C₆H₅COOH: (6 × 12) + (6 × 1) + (2 × 16) = 72 + 6 + 32 = 122 g/mol **Step 5: Calculate mass of benzoic acid** Mass = Moles × Molar mass X = 0.5 mol × 122 g/mol = 61 g **Therefore, the answer is 61 g.**