A chalk sample exactly 17.52 gram HCl for complete reaction with all `CaCO_(3)` present in it. If the chalk sample is 72% pure, the mass of sample taken is

Explanation

`CaCO_(3) + 2HCl to CaCl_(2) + H_(2)O+CO_(2)`
Moles of HCl `=(17.52)/(36.5)`
Moles of `CaCo_(3)=(1)/(2)xx(17.52)/(36.5)`
Weight of `CaCO_(3)` required `=(1)/(2)xx(17.52)/(36.5)xx100`
Mass of sample taken:
`=(1)/(2) xx(17.52)/(36.5)xx(100xx100)/(72)=33.33` gm