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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. |
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Answer» Solution :`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 % |
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