[dagitt]

^{this question has so much info and it's just confusing to me. any explanations or help would be great. i never learned about 'weight percentages' in class, is it a synonym for something else? anyway, here is the question...sorry if it is a little long!!}

Suppose you wish to analyze an iron ore for its content. In this case, the iron in the ore can be converted quantitatively to the iron (II) ion:

MnO₄^-1 + 5Fe⁺² + 8H⁺ --> Mn⁺² + 5Fe⁺³ + 4H₂O

MnO₄^-1 = purple, oxidizing agent
5Fe⁺² = colorless, reducing agent
8H⁺ = colorless
Mn⁺² = colorless
5Fe⁺³ = pale yellow


The MnO₄^- ion is a deep purple, but when reacts with the Fe⁺², the color disappears b/c the reaction product, the Mn⁺² ion is colorless. Thus, as KMnO₄ is added from a buret, the purple color disappears as the solutions mix. When all the Fe⁺² has been converted to Fe⁺³, any additional KMnO₄ gives the solution a permanent purple color. Therefore KMnO₄ solution is added from the buret until the initially colorless, Fe⁺² containing solution just turns a faint purple color, the signal that the quivalence point has been reached. Let us assume that a 1.025g sample of iron containing ore requires 24.35mL of .0195M KMnO₄ to reach the equivalence point. What is the weight percent of iron in the ore?





^{P.S. Where did the KMnO₄ come from?? That wasn't in the equation...i'm so confused....}

[enahs]

How many mols of KMnO4 is that? That corresponds to how many mols of iron? Remember your stoichiometry and balanced equation.

What is the mass of that many mols of iron?
What percentage is that of your 1.025 g sample?

The KMnO4 comes from the salt of the Potassium Permanganate. The potassium does not play a role in the reaction, it is just a spectator ion.