Question

10 g sample of limestone produces 1.84 L of CO2 gas at S.T.P. Determine percentage purity of limestone.​

Answer 1

Answer:

75%

Explanation:

75 %

SOLUTION:-

10g of limestone i.e. CaCO3 contains= 100g/mole10g moles of CaCO3=0.1 moles of CaCO3

CaCO3⟶CaO+CO2

1 mole of CaCO3 produce 1 mole of CaO

Thus 0.1 moles of CaCO3 must produce 0.1 mole of CaO

10g of CaCO3 must produce 0.1×56=5.6g of CaO

But CaO produce is 4.2g

Pure product obtained is 4.2g from 10g of CaCO3

Product that obtain along with 1 m purity from 10g of CaCO3 is 5.6g

So, percentage purity= mass of impure substance obtainedmass of pure substance obtained×100

% purity= 5.64.2×100=75%