25.0 cm3 of an acidified solution containing Fe2+ ions was titrated against potassium manganate (VII) solution. 20.0cm3 of 0.050M potassium manganate (VII) was needed. Calculate the concentration of Fe2+ ions in the acidified solution.

Answer 1

Write the equation and balance it.
(If you mean potassium manganate that is K2MnO4. If you mean potassium permanganate that is KMnO4. I suspect you mean KMnO4 since yu write (VII).
Using the coefficients in the balanced equation, convert moles KMnO4 to mols Fe^+2. Then M Fe = moles Fe/L Fe.

Answer 2

.000025MFe ???