[dyingsoul]

A new question:

1.5 g of an iron wire was dissolved in excess dilute sulphuric acid and the solution was made up to 200 cm3. 25.0 cm3 of this solution needed 25.45 cm3 of a 0.0188 mol dm-3 potassium manganate (VII)solution for oxidation. Calculate the percentage of iron in the iron wire.

What I got (please correct me if I am wrong):
Fe2+ => Fe3+ + e-
MnO4- + 8H+ + 5e- => Mn2+ + 4H2O

MnO4- + 8H+ + 5Fe2+ => Mn2+ + 4H2O + 5Fe3+

No of moles of MnO4- = 25.45/1000 * 0.0188 = 0.00047846 mol
Mole ratio, MnO4- : Fe2+ = 1 : 5
No of moles of Fe2+ in 25.0 cm3 = 0.00047846 * 5 = 0.0023923 mol
No of moles of Fe2+ in 200 cm3 = 0.0023923/25.0 * 200 = 0.019138 mol
Mass of Fe2+ = 0.019138 ( 55.8 + 32.1 + 16.0 *4) = 2.9071 g

I'm stuck here already. Cos apparently my answer for Fe2+ is more than what is provided.

[AWK]

Mass of Fe2+ = 0.019138 ( 55.8 + 32.1 + 16.0 *4) = 2.9071 g

Question concerns mass of iron, not iron sulfate.