Question

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

Answer 1

`MnO_(4)+4"HCl"toMnCl_(2)+Cl_(2)+2H_(2)O`
`2KI+Cl_(2)to2KCl+I_(2)`
`I_(2)+2Na_(2)S_(2)O_(3)toNa_(2)S_(4)O_(6)+2NaI`
`m" mol of "Na_(2)S_(2)O_(3)=30xx0.1=3.0`
`m" mol of "I_(2)=(3.0)/(2)=1.5`
`m" mol of "`Cl_(2)=1.5`
`m" mol of "MnO_(2)=1.5xx10^(-3)xx87=0.1305g`
% of `MnO_(2)=(0.1305xx100)/(0.5)=26.1%`