Total charge required for the oxidation of two moles `Mn_3O_4` into `MnO_4^(2-)` in presence of alķaline medium is `2.0 xx 10^(x)`, x is

Answer

1

Explanation

`Mn_3O_4 overset(OH^(-))to MnO_4^(-2)`
` +(8)/(3) " "+6`
charge in oxidation state =` 6 -(8)/(3)`
For 3 Mn atoms `=3[6-(8)/(3)] = 10`
For 1 mole = 10
For 2 moles =` 2 xx 10 = 20 = 2 xx 10^(1) , x =1`