Analytical chemistry

The iron in a 0.6656 g ore sample was reduced quantitatively to the +2 state and then titrated with 26.753 g of KMnO4 solution. Calculate the percent Fe2O3 in the sample.

the mass i get for fe2o3 is too big, but i don't know how else to do this problem, someone please help :S

5Fe + Mno4 -> 5Fe + Mn
n(MnO4)=26.753/158=0.1693 mol
n(Fe)=5*n(MnO4)=0.8466139 mol

2Fe->Fe2O3
m(Fe2O3)=0.8466139*159.7*(1/2)=67.6 g

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  1. I don't see a molarity for KMnO4 anywhere. Did I just miss it?

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  2. That was all that was given

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  3. Sorry. That's GRAMS KMnO4 so I don't need the molarity. Let me have a few minutes.

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  4. It's ok. Thank you for the help.

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  5. I don't see anything wrong with what you've done. I suspect the problem should read 0.26753g KMnO4. However, is there anything wrong with reporting Fe2O3 greater than 100%. You wouldn't expect Fe to be greater than 100% in a sample but %Fe2O3 or %Fe3O4 might be greater than 100%. Right? I note that 0.26753 g KMnO4 would give about 0.47 g Fe in the sample and that might be about right but that converted to Fe2O3 is higher than 100%.

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  6. Maybe there was a mistake in the book, because i always get the result higher than 100%

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  7. I expect you're right.

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  8. heh....i feel stoopid...

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