3.0 g sample of KOCl and `CaOCl_(2)` is dissolved in water to prepare 100 mL solution, which requried 100 " mL of " 0.15 M acidified `K_(2)C_(2)O_(4)`. For the point. The clear solution is now treated with excess of `AgNO_(3)` solution which precipitates 2.87 g of `AgCl`. Calculate the mass percentage of KOCl and `CaOCl_(2)` in the mixture.

Explanation

(a). `2e^(-)+undersetunderset(x=+1)(x-2=-1)(overset(+1)(ClO^(ɵ))toundersetx=-1)(Cl^(ɵ))(n=2)`
(b). `2e^(-)+undersetnderset(2x=0)(2x-2=-2)(Cl_(2)O^(-2))tounderset(2x=-2)(Cl^(ɵ))(n=2)`
(c). `C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)(n=2)`
Let a and b millimoles of KOCl and `CaCOCl_(2)` are present in the mixture.
`m" Eq of "KOCl+m" Eq of "CaOCl_(2)=m" Eq of "K_(2)C_(2)O_(4)`
`2a+2b=100xx0.15xx2` (n-factor)
`therefore2a+2b=30` .(i)
Also millomoles of `Cl^(ɵ)` from `KOCl+` millimoles of `Cl^(ɵ)` from `CaOCl_(2)-=` millimoles of AgCl`
`a(lCl^(ɵ)` ions) `+2b(2Cl^(ɵ)` ions)`=(2.87)/(143.5)xx10^(3)`
`[Mw of AgCl=143.5]`
`thereforea+2b=20` (ii)
From equation (i) and (ii) `a=10,b=5`
`% of KOCl=(10xx10^(-3)xx90.5xx100)/(3)` `[Mw of KOCl=90.5]`
`=30.1%`
`% of CaOCl_(2)=(5xx10^(-3)xx127)/(3)xx100` `[Mw of CaOCl_(2)=127]`
`=21.1%`