A sample of `FeSO_(4)` and `FeC_(2)O_(4)` is dissolved in `H_(2)SO_(4)` . The complete oxidation of sample required 8/3 eq. of `KMnO_(4)`. After oxidation , the reaction mixture was reduced by Z. On again oxidation by `KMnO_(4) ` required `5/3` eq. The mole ratio of `FeSO_(4)` and `FeC_(2)O_(4)` is

Options

A. `3:7`
B. `2:3`
C. `7:3`
D. `3:5`

Answer

C

Explanation

Let milli moles of `FeSO_(4) = x`
Let milli moles of `FeC_(2)O_(4) = y`
m.eqts `(FeSO_(4) +FeC_(2)O_(4))` = .eqts of `KMnO_(4)`
`x +3y = 40 xx (1)/(15) = (8)/(3) rArr 3x +9y = 8` ...(1)
By zinc and `H_(2)SO_(4)` only `Fe^(3+)` ions again reduced to `Fe^(2+)` ions
milli equivalents `Fe^(2+)` ions formed again are oxidised by `KMnO_(4)`
`x +y = 25 xx (1)/(15) =(5)/(3) rArr 3x +3y = 5` .....(2)
solving (1) & (2) `x = (7)/(6),y = (1)/(2),x :y = 7:3`