JEE Main 2025 Chemistry question
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) Consider the above reaction, what mass of CaCl₂ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of CaCO₃? (Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol⁻¹, respectively)
Answer: (3)
Solution
HCl is limiting reagent. Moles CaCl2 = 0.25×0.76/2. Mass = 0.25×0.76/2 × 111 = 10.545 g.