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
heh....i feel stoopid...
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.
To calculate the percent Fe2O3 in the sample, you need to use stoichiometry and the molar masses of the compounds involved.
1. Start by finding the number of moles of KMnO4 used:
- Mass of KMnO4 used = 26.753 g
- Molar mass of KMnO4 = 158 g/mol
- Number of moles of KMnO4 = Mass of KMnO4 used / Molar mass of KMnO4 = 26.753 g / 158 g/mol = 0.1693 mol
2. Since the iron in the ore sample was reduced quantitatively to the +2 state, the number of moles of Fe would be the same as the number of moles of KMnO4 used. Therefore, the number of moles of Fe is 0.1693 mol.
3. Now, use the balanced chemical equation to calculate the number of moles of Fe2O3:
- The balanced equation is: 5Fe + MnO4 --> 5Fe + Mn
- From the equation, you can see that for every 1 mole of MnO4, you have 5 moles of Fe.
- Therefore, the number of moles of Fe2O3 = 5 * number of moles of Fe = 5 * 0.1693 mol = 0.8466 mol
4. Finally, calculate the mass of Fe2O3 in the sample:
- Molar mass of Fe2O3 = 159.7 g/mol
- Mass of Fe2O3 = Number of moles of Fe2O3 * Molar mass of Fe2O3 = 0.8466 mol * 159.7 g/mol = 135.24 g
5. To find the percentage of Fe2O3 in the sample, use the formula:
- Percent Fe2O3 = (Mass of Fe2O3 / Total mass of sample) * 100
- Total mass of sample = 0.6656 g (given in the question)
- Percent Fe2O3 = (135.24 g / 0.6656 g) * 100 = 20321.35%
Therefore, the percent Fe2O3 in the sample is approximately 20321.35%.