Difficult Chemistry A Level back titration help

Can anyone help me to explain how to solve the answer after working out how many mols reacted.

Ethanedioic acid, H2C2O4, is found in a number of plants and spices, including rhubarb, tea leaves and black pepper. The ‘gritty feeling’ experienced when drinking milk with a rhubarb dessert is caused by the precipitation of calcium ethanedioate.
In the body, ethanedioate ions combine with iron(II) ions to form a precipitate of iron(II) ethanedioate, FeC2O4. Some of the iron(II) ethanedioate remains in solution and is excreted in urine.

(a) The concentration of iron(II) ethanedioate in urine can be determined by back titration.
The iron(II) ethanedioate is oxidised by cerium(IV) ions, Ce4+.
Fe2+(aq) + Ce4+(aq) → Fe3+(aq) + Ce3+(aq)
C2O42− (aq) + 2Ce4+(aq) → 2CO2(g) + 2Ce3+(aq)

After this reaction, the excess cerium(IV) ions are titrated with a standard solution of sodium ethanedioate.
50.0 cm3 of 0.200 mol dm−3 cerium(IV) sulfate solution, Ce(SO4)2(aq), was added to 25.0 cm3 of a sample of urine. The excess cerium(IV) ions were reduced to cerium(III) ions by adding 30.00 cm3 of 0.100 mol dm−3 sodium ethanedioate.

Calculate the concentration, in mol dm−3, of the iron(II) ethanedioate in the sample of urine.
(5)

The correct answer is 0.0533 mol dm−3

Reply 1

username3249896

number of moles of excess cerium(IV) ions = 6 x 10^-3 mol
number of moles of cerium(IV) ions added = 0.01 mol
number of moles of cerium(IV) ions reacted = 4 x 10^-3 mol
number of moles of FeC2O4 = 1.33 x 10^-3 mol
[FeC2O4] in urine = 0.0533 mol dm^-3

Reply 2

temtem12

Thank you!!😁

Original post by BobbJo

Reply 3

Munrhe

I am really sorry but I do not understand this line, all else makes perfect sense. Can you please explain? Thanks number of moles of FeC2O4 = 1.33 x 10^-3 mol

Reply 4

username3249896

Original post by Munrhe

I am really sorry but I do not understand this line, all else makes perfect sense. Can you please explain? Thanks number of moles of FeC2O4 = 1.33 x 10^-3 mol

1 mol of FeC2O4 reacts with 3 mol of Ce⁴⁺
so the number of moles of FeC2O4 = 4 x 10^-3 / 3 = 1.33 x 10^-3 mol

Reply 5

Munrhe

Thank you so much. That is what I thought based on the answer but the equation supplied in the specimen paper suggested fe2 ce4 etc so looked like 1:1 ratio. Thus, I did not understand.

Reply 6

username3249896

Original post by Munrhe

Thank you so much. That is what I thought based on the answer but the equation supplied in the specimen paper suggested fe2 ce4 etc so looked like 1:1 ratio. Thus, I did not understand.

1 mol of FeC2O4 contains 1 mol of Fe2+ and 1 mol of C2O4^2-
1 mol of Fe2+ reacts with 1 mol of Ce4+
1 mol of C2O4^2-reacts with 2 mol of Ce4+
so 1 mol of FeC2O4 reacts with 1 + 2 = 3 mol of Ce4+