How much amount of CaCO3 in gram (having 50 percent purity) produce 0.56 liter of CO2 at STP on heating?

CaCO3→∆CaO+CO2To produce 0.56 liter or 0.5622.4×10=140 mole  Amount of CaCO3 required to produce 140 mole of CO2=140×100=52 gram Percentage purity is = 50% So, amount required =2×52=5 gram