`4` g of an impure sample of `CaCO_(3)` on treatment with excess `HCl` produces `0.88` g `CO_(2)`. What is per cent purity of `CaCO_(3)` sample ?

Explanation

`CaCO_(3)+2HClrarrCaCl_(2)+H_(2)O+CO_(2)`
`44` g `CO_(2)-=100` g `CaCO_(3)`
`0.88` g `CO_(2)-=(100xx0.88)/(44)= 2.0 g CaCO_(3)`
`:.` Percentage purity `=(2)/(4)xx100= 50%`