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A mixture of FeO and Fe(2)O(3) is compl...

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 treated with Zn dust which converted `Fe^(3+)` of the solution to `Fe^(2+)`.The `Fe^(2+)` required 1000 mL of 0.10 `MK_(2)Cr_(2)O_(7)` solution. Find out the weight % `Fe_(2)O_(3)` in the mixture.

Text Solution

Verified by Experts

The correct Answer is:
80.85
in eq. Of FeO = m.eq of `KMnO_(4)`
`=0.25 xx 5 xx 100`
n-mole of FeO(n=1) `=(0.25 xx 100 xx 5)/1 = 125`
Total i-eq. Or m-mole of `Fe^(2+)`
`=1000 xx 0.1 xx 6=600`
(From FeO and `Fe_(2)O_(3)` after reaction with Zn dust m-mole of `Fe^(2+)` from `Fe_(2)O_(3) = 600-125= 475`)
m-mole of `Fe_(2)O_(3) = 475/2`
Mass of FeO `=(125 xx (56 + 16))/1000 = 38 g`
`% Fe_(2)O_(3) = 38/(38+9) xx 100 = 80.85%`
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