8.4 gm of MCO₃ on heating liberates 2.24 lit of CO₂ at STP. The atomic weight of metal ‘M’ is 

MCO₃$\stackrel{∆}{\to }$MO + CO₂

mass of MO formed = 8.4 - 4.4 = 4 gm ( 2.24 lit of CO₂ at STP = 4.4 gm)

$\frac{8.4}{\mathrm{E}+30}=\frac{4}{\mathrm{E}+8}$

E = 12

At .wt = Valency X Eq.Wt = 2 X 12 = 24