Question

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

Answer 1

Hey dude your answer is

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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