posted by Gloria .
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
I don't see a molarity for KMnO4 anywhere. Did I just miss it?
That was all that was given
Sorry. That's GRAMS KMnO4 so I don't need the molarity. Let me have a few minutes.
It's ok. Thank you for the help.
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%.
Maybe there was a mistake in the book, because i always get the result higher than 100%
I expect you're right.