Question

6.5g of an impure sample of limestone liberates 2.2g of CO$x_{2}$ on strong heating. The percentage purity of CaCo$x_{3}$ in the sample is

Answer 1

amount of CO2 liberated = 2.2 gm
1 mole CO2 = 44 gm
therefore 2.2 gm CO2= 1/44*2.2 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