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