Transition Metals / a level depression

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wannagotouni
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#1
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#1
Heyy ! ( sorry for the very long exam question but if someone knows how to solve it plz tell me LMAO )

(d) Iron(II) ethanedioate dihydrate can be analysed by titration using potassium manganate(VII) in acidic solution. In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.
A 1.381 g sample of impure FeC2O4.2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution.
25.0 cm3 of this solution decolourised 22.35 cm3 of a 0.0193 mol dm–3 solution of potassium manganate(VII).
(i) Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate.

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

Fe2+ Fe3+ + e–

C2O42– 2CO2 + 2e–

(ii) Calculate the percentage by mass of FeC2O4.2H2O in the original sample.
(If you have been unable to answer part (d)(i) you may assume that three moles of manganate(VII) ions react with seven moles of iron(II) ethanedioate.
This is not the correct ratio.)

(っ °Д °っ≡(▔﹏▔)≡
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Arctic Kitten
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I don't get what you write as the equation at all, though ofc I can guess, but please be clear and at least proofread.

So for first part, how much e- does one FeC2O4.2H2O molecule release? This should be easy (literally just addition). Let's say it's x.
MnO4- need 5e- to react.
So to balance, you need x mol MnO4- for every 5 mol of FeC2O4.2H2O.

For second part, use 22.35 cm3 of a 0.0193 mol dm–3 solution of potassium manganate(VII) to calculate the mol of MnO4-
using ratio from the first part, you will get the mol of FeC2O4.2H2O in 25.0 cm3 of solution.
initial solution has 250 cm, so you need to multiply to get the correct mol.
With the correct mol, use it to find the mass, then find % mass.
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Kenn Scott
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#3
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(Original post by wannagotouni)
Heyy ! ( sorry for the very long exam question but if someone knows how to solve it plz tell me LMAO )

(d) Iron(II) ethanedioate dihydrate can be analysed by titration using potassium manganate(VII) in acidic solution. In this reaction, manganate(VII) ions oxidise iron(II) ions and ethanedioate ions.
A 1.381 g sample of impure FeC2O4.2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution.
25.0 cm3 of this solution decolourised 22.35 cm3 of a 0.0193 mol dm–3 solution of potassium manganate(VII).
(i) Use the half-equations given below to calculate the reacting ratio of moles of manganate(VII) ions to moles of iron(II) ethanedioate.

MnO4– + 8H+ + 5e– Mn2+ + 4H2O

Fe2+ Fe3+ + e–

C2O42– 2CO2 + 2e–

(ii) Calculate the percentage by mass of FeC2O4.2H2O in the original sample.
(If you have been unable to answer part (d)(i) you may assume that three moles of manganate(VII) ions react with seven moles of iron(II) ethanedioate.
This is not the correct ratio.)

(っ °Д °っ≡(▔﹏▔)≡
Then calculate the percentage by mass of FeC2O4.2H2O in the original impure sample.

First, calculate the reacting molar ratio:

Each iron(II) ion provides one electron on oxidation and each ethanedioate ion provides two electrons: a total of three moles of electrons per iron(II)ethanedioate.

5 moles iron(II)ethanedioate provide 15 electrons and these reduce three moles manganate(VII) ions.

So the overall equation has to be:

5FeC2O4 + 3MnO4— + 24H+ ⟶ 3Mn2+ + 10CO2 + 5Fe3+ + 12H2O

Now you have the reacting ratio of manganate(VII) to iron(II)ethanedioate as 3 moles to 5 moles.

Second, you calculate the moles of manganate(VII) actually used in the titration and work back from this value.

Use n=cV

Moles MnO4— = 0.0193 × 22.35/1000 = 0.00043 moles

Third calculate the number of moles of the iron(II) salt in the 25ml of its solution:

Moles FeC2O4 = 0.00043 × 5/3 = 0.00072 moles

Fourth, note that 25ml sample was a tenth of the total sample dissolved in the 250ml volumetric flask. Therefore:

Total moles FeC2O4 = 0.0072moles

Fifth, calculate the molar mass of the iron(II) salt:
Fe— 55.8
C—12 × 2 = 24
O—16 × 4 = 64 +
H2O—18 × 2= 36

Giving a total Mr (FeC2O4 ) of 179.8 g.mol—1

Sixth, calculate the mass of pure iron(II) salt using mass = molar mass × moles

Mass of FeC2O4 = 179.8 × 0.0072 = 1.295g

Lastly, calculate the percentage by mass of the pure iron(II) salt since

% purity = pure mass × 100 /total mass

% purity = 1.295 × 100/1.381 = 93.7%
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