`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

Adding equation (i), (ii) and (iii) we get
`underline(K_(2)MnF_(6)+2SbF_(5)to2KSbF_(6)+MnF_(3)+(1)/(2)F_(2))` ..(iv)
In equation (iii)
`e^(-)+overset(+4)(Mn)tooverset(+3)(Mn)` (n factor`=1`)
`F^(ɵ)to(1)/(2)F_(2)+e^(-)` (n-factor`=1`)
Therefore according to equation (iv)
`2 " mol of "SbF_(5)-=1" mol of "K_(2)MnF_(6)`
`-=1 " Eq of "K_(2)MnF_(6)`
`therefore 1 " mol of "SbF_(6)=(1)/(2)Eq=0.5" Eq of "K_(2)MnF_(6)`