A sample of a mixture of `CaCl_(2)` and NaCl weighing 4 . 22 g was treated to precipitate all Ca as `CaCO_(3)` which was then heated and quantitatively converted to `0 . 959` g of CaO .Calculate the percentage of `CaCl_(2)` in the mixture. `(Ca = 40 , O = 16 , C = 12 and Cl = 35 . 5)`

Explanation

We have,
`{:(CaCl_(2) to CaCO_(3) to CAO),(" x g (say) 0 . 959 g" ):}`
Since Ca atoms are conserved, applying POAC for Ca atoms,
moles of Ca in ` CaCl_(2)` = moles of Ca in CaO
`1 xx ` moles of `CaCl_(2) = 1 xx ` moles of CaO
`(x)/( 111) = (0.959)/( 56) [{:(CaCl_(2) = 111),(" "CaO = 56 ):}] `
x = 1.901
Percentage of `CaCl_(2) = 1 . 901 xx (100)/( 4 . 22 = 45 . 04%`