`100 g CaCO_(3)` reacts with `1 litre 1 N HCl`. On completion of reaction how much weight of `CO_(2)` will be obtain

Options

A. `5.5g`
B. `11g`
C. `22g`
D. `33g`

Answer

C

Explanation

1L of 1N HCl = 1L of 1M HCl = 1mol of HCl
Amount of `CaCO_(3) = 100 g = 1 mol`
`underset(1mol)(CaCO_(3))+underset(2mol)(2HCl)toCaCl_(2)+underset(1mol)underset(44g)(CO_(2))+H_(2)O`
Thus HCl is the limiting reactant
2 mol of HCl after reaction with `CaCO_(3)` give `CO_(2) =` 1 mol
1 mol of HCl after reaction with `CaCO_(3)`, give `CO_(2) = (1)/(2)mol`
`:.` Mass of `CO_(2)`obtained `= (1)/(2)xx44 = 22g`