how much amount of CaCO3 in gram having percentage purity 50% produces 0.56l of CO2 at STP on heating?

how much amount of CaCO3 in gram having percentage purity 50% produces

1.	how much amount of CaCO3 in gram having percentage purity 50% produces 0.56l of CO2 at STP on heating?
Answer» Hey dude your ANSWER is ============================= At STP, any gas has volume 22.4L for 1mol. So, 0.56L of CO2 = 1×0.5622.4=0.025 mol Now, the reaction is: CaCO3 ⇌ CaO+CO2Reaction stoichiometryCompound Coefficient Molar Mass Moles WeightCaCO3 1 100 0.025 2.50CaO 1 56 0.025 1.40CO2 1 44 0.025 1.10 So, the real amount of CaCO3 = 2.5 g But, since the SAMPLE is 50% pure, so actual amount of CaCO3 = 5g Hope it helps you plzz MARK as brainliest