For the reaction
`M^(x+)+MnO_(4)^(ө)rarrMO_(3)^(ө)+Mn^(2+)+(1//2)O_(2)`
if `1 "mol of" MnO_(4)^(ө)` oxidises `1.67 "mol of" M^(x+) "to" MO_(3)^(ө)`, then the value of `x` in the reaction is

Answer

2

Explanation

`Mn^(7+) + 5e rarr Mn^(2+)`
`:. 1` mole of `MnO_(4)^(-)` accepts `5` moles of electrons.
`:. 5` mole of electron are lost by `1.67` mole of `M^(x+)`
`:. 1` mole of `M^(x+)` will lose electron `= (5)/(165) ~~ 3` moles
Since `M^(x+)` change to `MO_(3)^(-)` (oxidation no. of `M = +5`) by accepting `3` electron
`:. x = +5 - 3=+2`