The number of mole of `KMnO_(4)` that will be needed to react completely with one mole of ferrous oxalate in acidic solution is:

Options

A. 2/5
B. 3/5
C. 4/5
D. 1

Answer

B

Explanation

(b) `KMnO_4, + FeC_2O_4 rarr Fe^(3+) + 2CO_2, + Mn^(2+)`
So half reaction,
`MnO_(4)^(-) rarr Mn^(2+)` (decrease in O.N. = 5)
`FeC_2O_4 rarr Fe^(3+) +2CO_2` (increasc in O.N. = 1)
Now equating the change in O.N. and then by adding both half reactions we get `5FeC_2O_4 + MnO_(4)^(-) rarr 5Fe^(3+) +10CO_2 +Mn^(2+)`
on balancing equation,
`3MnO_(4)^(-) +5FeC_2O_4 + 24H^(+) rarr 5Fe^(3+) + 3Mn2^(2+) + 10CO_(2) + 12H_2O`
`therefore` 3 moles of `KMnO_4 = 5` moles of `FeC_2O_4`
I mole `FeC_2O_4` = (3/5) moles of `KMnO_4`