`H_(2)O_(2)` acts as both oxidising and reducing agent. As oxidising agent, its product is `H_(2)O` but as redusing agent, its product is `O_(2)`. Volume strength has great significance for chemical reactions. The strength of '10 V means 1 volume (or litre) of `H_(2)O_(2)` on decomposition `(H_(2)O_(2) rarr H_(2)O+1/2 O_(2))` gives 10 volumes (or litre) of oxygen at NTP.
`15 g Ba(MnO_(4))_(2)` sample containing inert impurity is completely reacting with 100 mL of '11.2 V' `H_(2)O_(2)`, then what will be the `%` purity of `Ba(MnO_(4))_(2)` in the sample: (Atomic mass: `Ba=137, Mn=55)`

Options

A. `50%`
B. `40%`
C. `68%`
D. `25%`

Answer

C

Explanation

m.eq of `Ba(MnO_(4))_(2)` reacted =m eq of `H_(2)O_(2)` reacted
`(w xx 1000)/(37.5) = (56xx 100)/(5.6)`
`:.` wt. of `Ba(MnO_(4))_(2) = 37.5g` the % purity of `Ba(MnO_(4))_(2)`
`(37.5)/(5%) xx 100 = 68.18%`