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

Options

A. `5`
B. `3`
C. `2`
D. `1`

Answer

C

Explanation

`5e^(-)+MnO_(4)^(ө)rarrMn^(2+)`
`M^(x+)rarrMO_(3)^(ө)+(5-x)`
`x-6 = -1`
`x=5`
`"Equivalent of" MnO_(4)^(ө) = "Equivalent of" M^(x+)`
`1 mol xx 5= 1.67 mol xx (5-x)`
`:. X=2`