NEET 2022 Chemistry Neutralization Reactions MCQ Question
What mass of 95% pure CaCO₃ will be required to neutralise 50 mL of 0.5 M HCl solution according to the following reaction?
CaCO₃ + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + 2H₂O(l) [Calculate upto second place of decimal point]
3.65 g
9.50 g
1.25 g
1.32 g
Correct Answer
Detailed Explanation
The reaction requires 1 mole of CaCO₃ to neutralize 2 moles of HCl. The moles of HCl are 0.025, so 0.0125 moles of CaCO₃ are needed. This corresponds to 1.32 g of 95% pure CaCO₃.
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