I'm stuck on the example transition metal redox titration on page 221 of the OCR textbook.
The question is; 2.5g impure iron is reacted with sulfuric acid up to 250cm3.
25cm3 of this was titrated against 24.8cm3 of 0.018moldm3 KMnO4. Calculate the percentage by mass of iron in the sample.
I have my mole ratio as 1:2, because Fe2+ has oxidation of 2+ and MnO4- as an oxidation as -1, so I need 2 moles of the latter?
0.018*0.0248= 0.000464 moles of MnO4-
/2 = 0.000232 Moles of Fe2+ in 25cm3
*10 = 0.00232 Moles Fe2+ in 250cm3
* 55.8 = 0.125g Fe in original sample.
(0.125g/2.5g)*100 = 5% by mass.
The correct answer is just under 50% by mass, so I'm out by a magnitude of 10. I can't think of any possible place I could justify x10, unless I'm converting cm3 to dm3 wrong.
Any help would be greatly appreciated
(OCR F325) Tricky Titration Watch
- Thread Starter
- 20-05-2016 20:48
- 20-05-2016 20:59
You need to use this equation
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
The mole ratio is 5:1
So you need to multiply moles of MnO4- by 5, and then by 10
- 21-05-2016 02:42
In addition to the answer above, the point you need to take away is that the Mn is doing the oxidation, being reduced from +7 in MnO4 to +2.