Correct Answer - B
`underset(1mol)(MCO_(3))overset(Delta" ")(to)MO+underset("22400 ml at STP")underset(1mol)(CO_(2))`
448 of `CO_(2)` is given by the metal carbonate = 2g
22400 of `CO_(2)` is given by the metal carbonate
`(2)/(448)xx22400g=100g`
`:.` Mol mass of `MCO_(3)=100`
M+60=100 or atomic mass of metal = 100-60=40
Eq. mass of metal = 40/2=20