Let the metal is M, so the formula of its carbonate is MCO3. Molar mass of MCO3=x+12+3×16 =(x+60)g/mol (Let atomic mass of M is x .) MCO3
Δ
→
MO+CO2 1 mol    1 mol = 22.4L =22400 mL ∵‌448cc‌‌(448mL)‌‌CO2 is produced from carbonate =2g. ∴22400cc‌‌CO2 will be obtained from carbonate =‌
2×22400
448
=100g ∴‌100=x+60 x=100−60=40g/mol Eq. wt. of metal =‌