1.249 g of a sample of pure `BaCO_(3)` and impure `CaCO_(3)` containing some CaO was treated with dil. HCl and it evolved 168ml of `CO_(2)` at NTP. From this solution, `BaCrO_(4)` was precipitated, filtered and washed. The precipitate was dissolved in dilute sulphuric acid and diluted to 100ml. 10ml of this solution, when treated with Kl solution, liberated iodine which required exactly 20ml of 0.05N `Na_(2) S_(2) O_(3)`. Calculate the percentage of CaO in the sample.

Explanation

`n_(CaCO_(3)) + n _(BaCO_(3)) = n_(CO_(2)) = ( 168)/( 22400) = 7.5 xx 10^(-3) ` .....(1)
`2BaCO_(3) rarr 2BaCrO_(4) overset( H^(+))(rarr) BaCr_(2) O_(7) overset( Kl) ( rarr) l_(2) -=Na_(2) S_(2) O_(3)`
eq. of `Na_(2) CO_(3)` = eq. of `I_(2) = ` eq of `BaCr_(2) O_(7) = ( 20 xx 10^(-3) xx 0.05 xx 100)/( 10) = 1 xx 10^(-12)`
Moles of `BaCr_(2) O_(7) = ( 1)/( 6) xx 10^(-2)`
Moles of `BaCrO_(4) = ( 2)/( 6) ( 1 xx 10^(-2))`
Moles of `BaCO_(3) = ( 1)/( 3) xx 10^(-2) = 3.33 xx 10^(-3) ` .....(2)
Weight of `BaCO_(3) = 0.650 g m`
From equation (1) and (2) we get
`n_(CaCO_(3)) = 4.17 xx 10^(-3)`
weight of `CaCO_(3) = 100 xx 4.17 xx 10^(-3)`
=0.417g
weigght of `CaO = 1.249 - 0.656 -0.417 = 0.176`
% of CaO `= ( 0.176 )/( 1.249) xx 100`
= 14.09 %