[S_100]

50 cm³ of 0.2 mol dm^-3  Ce⁴⁺ added to solution of 25cm³ of Fe²⁺,
excess Ce⁴⁺ reduced to Ce³⁺ by adding 30 cm³ of 0.1mol dm^-3  of C₂O₄^2-.

Calculate concentration of Fe²⁺ in its initial solution (25 cm³).  My Q is up to finding mol of Fe²⁺

Back titration Equation:
-------------------------------------
Reaction 1:
Fe²⁺ + Ce⁴⁺ -----> Fe³⁺ + Ce³⁺
Reaction 2:
C₂O₄^2- + 2Ce⁴⁺ ------->2 CO₂ + 2Ce³⁺
------------------------------------
Calculation

Fe²⁺ react with excess Ce⁴⁺,  and this leftover Ce⁴⁺ from this first reaction,  titrated against C₂O₄^2-

mol of titre C₂O₄^2-  = 0.003
thus mol of  leftover Ce⁴⁺ that reacts with this oxidising agent = 0.006 (2 x 0.003)

total mol of Ce⁴⁺ added to solution of Fe²⁺  is given as 0.01

thus initial mol of Ce⁴⁺ that react with Fe²⁺  is 0.01 - 0.006 = 0.004

 mol of Fe²⁺ =


What I am confused about is calculating mol of Fe²⁺. A markscheme indicates that the mol of Fe²⁺ = 1/3 x mol of  Ce⁴⁺  which reacts with it.

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

Fe²⁺ + 3Ce⁴⁺ +C₂O₄^2----> Fe³⁺ + 3Ce³⁺ + 2 CO₂



But I don't understand why this is necessary when we know from reaction 1 mol of Fe²⁺  is oxidised by 1 mol Ce⁴⁺, thus mol of Ce⁴⁺ which reacts in reaction 1 is also  = mol of Fe²⁺ from 25cm³


Why do we need the overall reaction equation to work out how many mols of Fe²⁺ initially added?

[chenbeier]

Your calculation is correct. The Statement that 1/3 Ce⁴⁺ = Fe²⁺ is also correct according the total equation,  but cannot be used, because we have a subtraction of the cerium solution.
You dont know  how much  Cerium will be used for each in advance.
You have 0,01 mol cerium total, but can not say its one third of iron.