In qualitative estimation of any element using an oxidising agent is essential to predict which substance (s) gets oxidised by the oxidising agent. In case more than one substance is getting oxidised then the oxidising agent gets distributed in all the reactions taking place. From this information and the data given below answer the following questions:
`Fe^(2+) rarr Fe^(3+) +e^(-)`
`E^(@) =- 0.77V`
`MnO_(4)^(-) +5e^(-) rarr Mn^(2+) + 4H_(2)O`
`E^(@) = + 1.51V`
`2Cl^(-) rarr Cl_(2) +2e^(-)`
`E^(@) =- 1.36V`
`2SO_(4)^(2-) rarr S_(2)O_(8)^(2-) +2e^(-)`
`E^(@) =- 2.0V`
`C_(2)O_(4)^(2-) rarr 2CO_(2) +2e^(-)`
`Cr_(2)O_(7)^(2-) +14H^(+) +6e^(-) rarr 2Cr^(3+) +7H_(2)O`
`E^(@) = + 1.33` volt
Millimol of `FeC_(2)O_(4)` in the solution if 50ml of `0.1M KMnO_(4)` is used for its oxidation in presence dilute `HCl` if 3.5 millimoles of `Cl_(2)` is obtained along with other products.

Options

A. `(25)/(3)` m. mol
B. 6m. Mol
C. `12.5` m mol
D. Quanitative estimation can be made from the give data

Answer

B

Explanation

No of meq of `KMnO_(4) =` no of meq `Cl^(-)+` no of meq of `FeC_(2)O_(4)`
No of meq of `Cl^(-) =` no of meq of `Cl_(2) = 3.5 xx 2 = 7`
(n-factor =2)
`:.` no of meq of `FeC_(2)O_(4) = 25 - 7 = 18`
`:.` no of m mol of `FeC_(2)O_(4) = 18//3 = 6`