6.5g of an impure sample of limestone liberates 2.2g of CO on strong heating. The percentage purity of CaCo in the sample is

6.5g of an impure sample of limestone liberates 2.2g of CO on strong h

1.	6.5g of an impure sample of limestone liberates 2.2g of CO on strong heating. The percentage purity of CaCo in the sample is
Answer» Amount of CO2 liberated = 2.2 gm 1 MOLE CO2 = 44 gm therefore 2.2 gm CO2= 1/442.2 mole = 0.05 mole now CaCO3 = CaO +CO2 so from 1 mole CaCO3 we get 1 mole CO2 so to get 0.05 mole CO2 we NEED 0.05 mole CaCO3 mol. mass of CaCO3 = 100 1 mole CaCO3 = 100 gm 0.05 mole CaCO3 = 0.05 100 = 5 gm therefore purity of SAMPLE = 5/6.5*100 = 77% appx