Published by:
CGP EDU Academic Team
Published on: August 21, 2026
What mass of 95% pure CaC0 3 will be required to neutralise 50 mL of 0.5 M HC1 solution according to the following reaction?
CaCO 3(S ) + 2HCl (aq) 
CaCl 2(aq) + CO 2(g) + 2H 2 O (l)
[Calculate up to second place of decimal point]
Text Solution
Verified by ExpertsThe correct answer is:
Consider mass of CaCO 3 required as 'm' gram.
Pure CaCO 3 in gram = 
m
Mole of CaCO 3 = 
Moles of HCl required = 2 x moles of CaCO 3 =2x 
x 
m = 1.315g 
1.32g
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