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A sample of CaCO3 and MgCO3 weighed 2.21 gm is ignited to constant weight of 1.152 gm. What is the composition of the mixture? Also report the volume of CO2 produced.

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∴ (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

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