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

c) `5e^(-) + MnO_(4)^(-) rarr Mn^(2+)`
`M^(x+) rarr MO_(3)(-) + (5-x)`
`x + 6 xx (-2) = -1 therefore x=5` Equivalent of `MnO_(4)^(-)` = Equivalent of `M^(x+)` 1 mol `xx`=5 = 1.67 mol `xx` (5- x) 1 mol `xx` 5 1.67 mol (5- x) `therefore x = 2`