`F_(2)` can be prepared by reacting hexfluoro magnante (IV) with antimony pentafluoride as:
`K_(2)KnF_(6)+SbF_(5)overset(150^(@)C)toKSbF_(6)+MnF_(3)+F_(2)` ltBrgt The number of equivalent of `K_(2)MnF_(6)` requried to react completely with one " mol of "`SbF_(5)` in the given reaction is

Options

A. `1.52`
B. `5.0`
C. `0.5`
D. `4.0`

Answer

C

Explanation

`K_(2)MnF_(6) = 2SbF_(5) rarr 2KSbF_(6) +MnF_(3) +(1)/(2)F_(2)`
`overset(+4)(Mn)+e^(-) rarr overset(+3)(Mn)` (n-factor =1)
`F^(-) rarr (1)/(2)F_(2)+e^(-)` (n-factor =1)
`:.2` moles of `SbF_(5) -= 1` moles of `K_(2)MnF_(6) -=1` eqt of `K_(2)MnF_(6)`
`:. 1` moles of `SbF_(5) = (1)/(2)` eqt of `K_(2)MnF_(6)`