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

Then 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_4Fe(CN)_6) = 61/10`, so Moles of `Ba(MnO_4)_2 = 61/10 xx 1 = 6.1`