1.0 g a of moist sample of a mixture of KCl and `KClO_3` was dissolved in water and made up to `250mL`. 25 " mL of " this solution was treated with `SO_2`. The chlorate was reduced to chloride and excess of `SO_2` was removed by boiling. The total chloride was precipitated as `AgCl` . The weight of the precipitate was `0.1435 g`. In another experiment, 25 " mL of " the original solution was heated with 30 " mL of " 0.2 N solution of ferrous sulphate, and the unreacted ferrous sulphate required 37.5 " mL of " 0.08 N solution of an oxidising agent for complete oxidation. Calculate the molar ratio of the chlorate to the chloride in the given mixture `Fe^(2+)` reacts with `ClO_3^(ɵ)` according to the equation.
`ClO_3^(ɵ)+6Fe^(2+)+6H^(o+)toCl^(ɵ)+6Fe^(3+)+3H_2O`

Explanation

`ClO_(3)^(-)` is reduced to `Cl^(-) " by " SO_(2) " and " ClO_(3)^(-)` is also reduced to `Cl^(-) " by " Fe^(2+) ` , hence AgCl is formed due to total `Cl^(-)`
Meq. of `Fe^(2+) ` initially taken ` = 30 xx 0.2 = 6`
Meq. of `Fe^(2+)` unused ` = 37 . 5 xx 0.08 = 3`
` :. ` Meq of `Fe^(2+) " used " = 6.0 - 3.0 = 3.0`
Thus, Meq. of `ClO_(3)^(-) " in " 25 ml = 3.0`
Moles of `ClO_(3)^(-) " in " 25 ml = (3.0)/(1000 xx 6) = 0.0005`
`.^(+5)ClO_(3)^(-) to .^(-1)Cl^(-) ` (n-factor 6)
0.N `5 .... -1`
Thus , moles of `ClO_(3)^(-)` in 25 ml solution ` = 0.0005`
`ClO_(3)^(-)` is also reduced to `Cl^(-)` by `SO_(2)` in first experiment and precipitated as AgCl.
Thus , `Cl^(-)` formed from `ClO_(3)^(-) = AgCl` from `ClO_(3)^(-) = 0.0005`
Total AgCl formed both from actual and `Cl^(-) " from " ClO_(3)^(-) = 0.1435` gm
` = (0.1435)/(143.5) = 0.0010` mol
Thus, AgCl formed due to `Cl^(-) ` only = `0.0010 - 0.0005 = 0.0005` mol
Thus , `ClO_(3)^(-) " and " Cl^(-)` are in molar ratio ` = 1 : 1 `.