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

Total `Cl^(ɵ)` (from `KCl+KClO_3)(ClO_3^(ɵ)toCl^(ɵ))`
25 " mL of " the mixture`=0.1435 g of AgCl`
`=(0.1435)/(143.5)` " mol of "AgCl
`=10^(-3) " mol of "Cl^(ɵ)` ions
Second experiment:
Total volume of `FeSO_4=30mL`
Excess volume of `FeSO_4=(37.5xx0.08)/(2)=15mL`
Volume of `FeSO_4` used `=30-15=15mL`
`15 mL` of `0.2N FeSO_4` used up
`=15xx0.2xx10^(-3) " Eq of "FeSO_4`
`=3xx10^(-3) " mol of "FeSO_4`
Since 1 mol `ClO_3^(ɵ)` uses 6 " mol of "`FeSO_4`,
`3xx10^(-3)` " mol of "`FeSO_4=(1)/(6)xx3xx10^(-3)`
`=0.5xx10^(-3) " mol of "ClO_3^(ɵ)`
`Cl^(ɵ)+ClO_3^(ɵ)=10^(-3)`
`thereforeCl^(ɵ)=10^(-3)-0.5xx10^(-3)=0.5xx10^(-3)`
`thereforeCl^(ɵ)=0.5xx10^(-3),ClO_3^(ɵ)=0.5xx10^(-3)`