Question

150 mL of `N//10` HCl is required to react completely with 1.0 g of a sample of limestone. Calculate the percentage purity of calcium carbonate.

Answer 1

`150 mL (N)/(10)HCl-=150 mL (N)/(10)CaCO_(3)`
`underset("Mol.mass")(CaCO_(3))+underset(2g eq.)(2HCl) to CaCl_(2)+H_(2)O+CO_(2)`
Eq. mass of `CaCO_(3)=(40+12+48)/(2)=(100)/(2)=50`
Mass of `CaCO_(3)` present in 150 mL `N//10` solution.
`[NxxExx(V)/(1000)]=50xx(1)/(10)xx(150)/(1000)=0.75g`
Purity `=(0.75)/(1)xx100=75%`