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.