What mass of 95 \% pure {CaCO}_{3} will be required to neutralise 50 {mL} of 0.5 {M} {HCl} solution according to the following reaction? {CaCO}_{3( {s})}+2 {HCl}_{\text{(aq) }} \to {CaCl}_{2(\text{ aq })}+ {CO}_{2(\text{ g })}+2 {H}_{2} {O}_{\text{(l) }} [Calculate upto second place of decimal point] [NEET-2022]

Class 11 NEET Chemistry Some Basic Concepts Of Chemistry

Question : 12 of 63

Marks: +1, -0

What mass of 95% pure CaCO3 will be required to neutralise 50mL of 0.5MHCl solution according to the following reaction?
CaCO3(s)+2HCl‌(aq) ‌→CaCl2(‌ aq ‌)+CO2(‌ g ‌)+2H2O‌(l) ‌
[Calculate upto second place of decimal point]

[NEET-2022]

1.25g
1.32g
3.65g
9.50g

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