A 0.5 g sample containing `MnO_(2)` is treated with HCl liberating `Cl_(2)`. The `Cl_(2)` is passed into a solution of KI and `30.0 cm^(3)` of 0.1 M `Na_(2)S_(2)O_(3)` are required to tirate the liberated iodine. Calculate the percentage of `MnO_(2)` in the sample.

Explanation

`30.0 mL 0.1 M Na_(2)S_(2)O_(3)-=30.0 mL 0.1 N Na_(2)S_(2)O_(3)`
`-=30.0 mL 0.1 N I_(2)`
`-=30.0 mL 0.1 N Cl_(2)`
`-=30.0 mL 0.1 N MnO_(2)`
Amount of `MnO_(2)` present `=(NxxExxV)/(1000)`
`=(1)/(10)xx(87)/(2)xx(30)/(1000)`
`% MnO_(2)=(87xx30xx100)/(10xx2xx1000xx0.5)=26.1`