A mixture of `FeO` and `Fe_(2)O_(3)` is completely reacted with 100 mL of 0.25 M acidified `KMnO_(4)` solution. The resultant solution was then titrated with Zn dust which converted `Fe^(3+)` of the solution to `Fe^(2+)`. The `Fe^(2+)` required 1000 mL of `0.10 M K_(2)Cr_(2)O_(7)` solution. Find out the weight % `Fe_(2)O_(3)` in the mixture.

Options

A. 80.85
B. 19.15
C. 50
D. 89.41

Answer

A

Explanation

meqs of FeO= meqs of `KMnO_(4)=0.25xx5xx100`
meqs of `FeO(n=1)=(0.25xx100xx5)/(1)`
`[overset(+2)(2FeO)tooverset(+3)(Fe_(2)O_(3))+2e^(-)]=125`
Total meqs of `Fe^(2+)`
`=1000xx0.1xx6=600`
(From FeO and `Fe_(2)O_(3)` after reaction with Zn dust)
meqs of `Fe^(2+)` from `Fe_(2)O_(3)=600-125`, mmol of `FeO=(125)/(1)=125` mmol
=475
mmole of `Fe_(2)O_(3)=(475)/(2)` w.t of `FeO=(125xx(56+16))/(1000)g=9g`
wt. of `Fe_(2)O_(3)=(475)/(2)xx(160)/(1000)=38g`
`%Fe_(2)O_(3)=(38)/(38+9)xx100=80.85%`.