Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified `KMnO_(4)` for complete oxidation ?

Options

A. `FeSO_4`
B. `FeSO_3`
C. `FeC_2O_4`
D. `Fe(NO_2)_2`

Answer

A

Explanation

`2KMnO_4 + 3H_2SO_4 to K_2SO_4 + 2MnSO_4 + 3H_2O + 5[O]…(i)`
(a) `[2FeSO_4 + H_2SO_4 + [O] to Fe_2 (SO_4)_3 + H_2O]5…(ii)`
10 moles of `FeSO_4` requires = 2 moles of `KMnO_4`
or 1 mole of `FeSO_4` requires = 1/5 mole of `KMnO_4`
(b) `[2FeSO_4 + H_2SO_4 + 3[O] to Fe_2 (SO_4)_3 + H_2O]5 ...(iii)`
By multiplying eq. (i) `xx3` and eq. (iii) `xx 5` and adding we get
10 moles of `FeSO_3` requires = 6 moles of `KMnO_4`
1 moles of `FeSO_3` requires = 3/5 moles of `KMnO_4`
(c) `[2FeC_2O_4 + 3H_2SO_4 + 3[O] to Fe_2(SO_4)_3 + 3H_2O +4CO_2]5...(iv)`
By multiplying eq. (i) `xx3` and eq. (iv) `xx5` and adding we get
10 moles of `FeC_2O_4` requires = 6 moles of `KMnO_4 `
`1 mole of `FeC_2O_4` requires = 3/5 moles of `KMnO_4`
(d) `2Fe(NO_2)_2 + 3H_2SO_4 + 5[O] to Fe_2(SO_4)_3 + 4HNO_3 + H_2O ...(v)`
By adding eq. (i) and (v) we get
2 moles of `Fe(NO_2)_2` requires = 2moles of `KMnO_4`
1 mole of `Fe(NO_2)_2` requires least amount of `KMnO_4` for complete oxidation.