How Do You Calculate the Mass Percent of Iron in an Ore Using Redox Titration?

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To calculate the mass percent of iron in an ore using redox titration, the iron in the ore must first be reduced to Fe 2+ ions in an acidic solution. The titration is performed with a standard potassium permanganate solution, where the endpoint is indicated by a faint pink color due to the presence of MnO4-. A sample weighing 0.3500g is titrated with a 0.01621 M KMnO4 solution, using 41.56 mL to reach the endpoint. The discussion emphasizes the importance of correctly identifying half-reactions for both oxidation and reduction processes, ensuring that electrons can be canceled out to form a balanced redox equation. Understanding these reactions is crucial for accurately determining the mass percent of iron in the sample.
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Is anyone here able to help me or point me in the right direction?

Iron ores often involve a mixture of oxides and contain both Fe 2+ and Fe 3+ ions. Such an ore may be analyzed for its iron content by dissolving it in an acidic solution, reducing all of the iron to Fe 2+ ions and then titrating with a standard solution of potassium permanganate until there is just a tinge of pingk color in the solution containing the iron ions. (hint: the permanganate ion is intensely purple and Mn 2+ is colorless.) After the titration was complete, permanganate would be reduced and Fe 2+ would be oxidized.

A sample of iron ore weighing 0.3500g was dissolved in acidic solution, and all of the iron was reduced. The resulting solution was titrated with a 1.621x10^-2 Molar KMnO4 Solution. The titration endpoint was reached when 41.56 mL of the KMnO4 solution was used. Determine the mass percent of iron in the iron ore sample.

What I am thinking is that I need to create two half reactions with FeO and Fe2O3 --> Fe 2+. So I go through that and I end up with half reactions of :

2H+ + FeO -> Fe2+ + H2O
2e- + 6H+ + Fe2O3 -> 2Fe2+ + 3H2O

and then combine it to form

8H+ + 2e- + FeO + Fe2O3 --> 3Fe2+ + 4H2O

From what I've learned you are supposed to be able to cancel out the electrons when you combine the two half reactions, however in this case I only have electrons on one of the half reactions and thus I have no hope of canceling them out. Am I thinking about this correctly or is there something else I should be doing?
 
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Your thinking is on the right track, but you are only considering one of the half-reactions there.

A sample of iron ore weighing 0.3500g was dissolved in acidic solution, and all of the iron was reduced.
I think the question’s wording here is a little confusing, but let's assume that it means all of the Iron ions in solution have been converted (reduced) to Fe+2 (aq), and not all the way back to Iron metal (which is another, however illogical, interpretation of that statement).
So you have a solution of Fe+2 (aq)

When finding the half-reactions, don’t worry about the reaction which occurs to dissolve the Iron Oxide, just assume that the Iron starts out already in solution in the form of Fe+2 ions.

The half-reactions we are talking about here make up a RedOx (Reduction-Oxidation) reaction.
One half reaction will be the reduction portion ("Gain electrons, Reduction"), the other half reaction will be the oxidization portion (Loose electrons, Oxidization).

What is oxidized in the reaction?
What is reduced?
Put these two reactions together to get the complete RedOx reaction, and the electrons should/will cancel out with the proper coefficients.
 
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