## Question

### Solution

Correct option is

22 g

CaCO3  +  2 HCl → CaCl2 + H2O + CO2

100 g        2(36.5)                             44 g

= 73 g

1 L of 1 N HCl contains 1 g eq. of HCl, viz, 36.5 g.

Hence, HCl is the limiting reactant. As 73 g HCl

liberate CO2 = 44 g, therefore, 36.5 g HCl will liberate CO2 = 22 g.

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