In redox reaction, `Ba(MnO_4 )_2` oxidizes `K_4[ Fe(CN)_6 ]` into `Fe^(3+), CO_(3)^(2-) and NO_(3)^(-)` ions in acidic medium, where `Ba(MnO_4)_2` itself reduces into `Mn^(+2)` then how many moles of `Ba(MnO_4)_2` will react with 1 mole of `K_4 [Fe(CN)_6 ]`

Options

A. `61`
B. `1.6`
C. `6.1`
D. `61.1`

Answer

C

Explanation

The n-factor of `Be(MnO_4)_2` = 10 and n-factor of `K_4 [Fe(CN)_6].`=61. Now, since `("Moles of" Ba(MnO_4)_2)/(1 mol "of "K_4 Fe(CN)_6) = (61 )/(10)` so Moles of `Ba (MnO_4)_2 = (61)/(10 ) xx 1 = 6.1`