`10 g` of a sample of a mixture of `CaCl_(2)` and `NaCl` is treated to precipitate all the calcium as `CaCO_(3)`. This `CaCO_(3)` is heated to convert all the `Ca` to `CaO` and the final mass of `CaO` is 1.62 g.The percent by mass of `CaCl_(2)` in the origial mixture is

Explanation

`underset(1 mol)(CaCl_(2))tounderset(1 mol)(CaCO_(3))tounderset(1 mol)(CaO)`
Let x g of `CaCl_(2)`
Mole of `CaCl_(2)=(x)/(111)`
Mole of `CaO=(1.62)/(56)`
`therefore(x)/(111)=(1.62)/(56)becausex=3.21g`
`% of CaCl_(2)=(3.21)/(10)xx100=32.1%`