∴ (x + y) = 2.21 gm …(1)
∴ 100 gm of CaCO3 gives 56 gm of CaO
∴ x gm of CaCO3 ≡ 56 x / 100 g of CaO
Similarly 84 gm of MgCO3 gives 40 gm of MgO
∴ y gm of MgCO3 = 40 y / 84 gm of MgO
∴ Wt. of residue = 56 x / 100 + 40 y / 84 = 1.152 …(2)
Solving equations (1) and (2)
x = 1.19 gm; y = 1.02 gm
Mole of CO2 formed = Mole of CaCO3 + Mole of MgCO3
= 1.19 / 100 + 1.02 / 84 = 0.0241
∴ Volume of CO2 at STP = 0.0421 × 22.4 litre = 539.8 ml