0.05g of a commercial sample of `KClO_(3)` on decomposition liberated just sufficeint oxygen for complete oxidation of 20 mL CO at `27^(@)C` & 750 mm pressure. Calculate & of `KClO_(3)` in the sample.

Explanation

`n_(CO)=(PV)/(RT)=(750xx20)/(760xx0.0821xx300xx1000)=8.01xx10^(-4)`
`CO+(1)/(2)O_(2) to CO_(2),n_(CO)//n_(O_(2))=2`
`therefore` Moles of `O_(2)` required `(n_(O_(2)))=(n_(CO))/(2)=(8.01xx10^(-4))/(2)`
Reaction involved: `2KClO_(3) to 2KCl+3O_(2)`
3 mol `O_(2)` is produced by 2 mol `KClO_(3)`.
`thereforeO_(2)` formed`=(2)/(3)xx(8.01xx10^(-4))/(2)=2.66xx10^(-4)mol`
`therefore`Weight of `KClO_(3)=2.66xx10^(-4)xx122.5g=3.27xx10^(-2)g`
`therefore%` of `KClO_(3)` in the mixture`=(3.24xx10^(-2))/(0.5)xx100=65.4`