Determining Iron Oxide Composition in a Sample using Redox Titration

[steven10137]

Homework Statement

A sample of iron ore consisting of a mixture of FeO and Fe2O3 was dissolved in dilute sulfuric acid. The resultant solution was divided into two equal aliquots. The first aliquot was titrated with a potassium permanganate solution containing 6.30g of KMnO4 per litre, and required 15.0mL for complete reaction. The second aliquot was reduced with zinc and the solution then titrated with the potassium permanganate solution. 25.1mL was required for the second oxidation. Calculate the mass of each iron oxide in the original sample.

Homework Equations

n=m/M, n=cV

The Attempt at a Solution

1st aliquot:
Iron ore titrated with KMnO4
V(KMnO4) = 0.015 L
c(KMnO4) = 6.3 g/L
therefore: n(KMnO4) = cV = 6.3 x 0.015 = 0.0945 mol

2nd Aliquot:
Iron ore reduced with zinc and then titrated with same KMnO4
V(KMnO4) = 0.0251 L
c(KMnO4) = 6.3 g/L
therefore: n(KMnO4) = cV = 6.3 x 0.0251 = 0.15813 mol

I am totally confused and have no idea how to relate these values back in terms of the iron ore sample and how to even approach solving the problem.

Any Help would be greatly appreciated
Steven
by the way, this is an assignment out of my workbook and the answers (with no solutions) are:
m(FeO)=0.429 g
m(Fe2O3)=0.321 g

 

[chemisttree]

You need to write down the pertinent reactions that are occurring. You should know that potassium permanganate is a strong oxidizer and that Fe2O3 is the most oxidized form of iron in this example (potassium permanganate does not react with Fe2O3).

 

[Kushal]

first aliquot: only FeO will react. Fe2O3 has Fe3+...which cannot be further oxidised.

5Fe2+ + MnO4- + 8H+ ------> Mn2+ + 5Fe3+ + 4H2O

mass of KMnO4- required: (0.015*6.3)
mol of KMnO4- required: 0.0945\158 = 5.98*10^-4
1 mol MnO4- reacts with 5 mol Fe2+
5.98*10^-4 mol MnO4- reacts with 5*(5.98*10^-4) mol Fe2+
=3*10^-3 mol Fe2+
therefore number of moles of FeO in first aliquot is 3*10^-3 mol
mass of FeO in original sample is [(3*10^-3)*(56+16)]*2 = 0.429g

seocnd aliquot:

zinc will reduce only Fe2O3... FeO cannot obviously be reduced since it is in the lowest possible oxidation state.

Zn + 2Fe3+ -------> Zn2+ + 2Fe

this time there is the Fe2+ from the FeO and that from the reduced Fe2O3...
but the same equation as above can be used...

5Fe2+ + MnO4- + 8H+ ------> Mn2+ + 5Fe3+ + 4H2O

mass of KMnO4- required: (0.0251*6.3)
mol of KMnO4- required: 0.1581\158 = 0.001 mol
1 mol MnO4- reacts with 5 mol Fe2+
0.001 mol MnO4- reacts with 5*(0.001) mol Fe2+
=0.005 mol Fe2+
therefore number of moles of Fe2+ in first aliquot is 0.005 mol
it is already known that Fe2+ from FeO in one aliquot is 3*10^-3 mol
so, the number of moles of Fe2+ from the reduced Fe2O3 is (0.005-3*10^-3)
= 0.002mol

number of moles of Fe3+ in one aliquot = 0.002 using second equation.
since 1 mol Fe2O3 gives 2 mol Fe3+
number of mol of Fe2O3 in one aliquot is 0.002\2 = 0.001 mol
mass of Fe2O3 in original sample is [(0.001)*(56+56+16+16+16)]*2 = 0.321g

the values have been rounded off in the working...

Hope to have helped you

 

[steven10137]

Kushal u legend,

totally understand it now thanks a million!
also thanks chemistree; looking at Kushal's working, what you said before makes perfect sense, I was just unsure how to put that into context

cheers Steven