50 cm

^{3} of 0.2 mol dm

^{-3} Ce

^{4+} added to solution of 25cm

^{3} of Fe

^{2+},

excess Ce

^{4+} reduced to Ce

^{3+} by adding 30 cm

^{3} of 0.1mol dm

^{-3} of C

_{2}O

_{4}^{2-}.

Calculate concentration of Fe

^{2+} in its initial solution (25 cm

^{3}). My Q is up to finding mol of Fe

^{2+}Back titration Equation:

-------------------------------------

Reaction 1:

Fe

^{2+} + Ce

^{4+} -----> Fe

^{3+} + Ce

^{3+}Reaction 2:

C

_{2}O

_{4}^{2-} + 2Ce

^{4+} ------->2 CO

_{2} + 2Ce

^{3+}------------------------------------

Calculation

Fe

^{2+} react with excess Ce

^{4+}, and this leftover Ce

^{4+} from this first reaction, titrated against C

_{2}O

_{4}^{2-} mol of titre C

_{2}O

_{4}^{2-} = 0.003

thus mol of leftover Ce

^{4+} that reacts with this oxidising agent = 0.006 (2 x 0.003)

total mol of Ce

^{4+} added to solution of Fe

^{2+} is given as 0.01

thus initial mol of Ce

^{4+} that react with Fe

^{2+} is 0.01 - 0.006 = 0.004

mol of Fe

^{2+} =

What I am confused about is calculating mol of Fe

^{2+}. A markscheme indicates that the mol of Fe

^{2+} = 1/3 x mol of Ce

^{4+} which reacts with it.

I assumed they got this from the overall equation of the back titration

Fe

^{2+} + 3Ce

^{4+} +C

_{2}O

_{4}^{2-}---> Fe

^{3+} + 3Ce

^{3+} + 2 CO

_{2} But I don't understand why this is necessary when we know from reaction 1 mol of Fe

^{2+} is oxidised by 1 mol Ce

^{4+}, thus mol of Ce

^{4+} which reacts in reaction 1 is also = mol of Fe

^{2+} from 25cm

^{3} Why do we need the overall reaction equation to work out how many mols of Fe

^{2+} initially added?