Solving Redox Titrations: Na2SO3 & KMnO4

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To determine the volume of 0.0200 M KMnO4 needed to oxidize 40.0 mL of 0.100 M FeSO4 in sulfuric acid, one must first balance the redox reaction between MnO4- and Fe2+. The balanced equation indicates that for every mole of KMnO4, five moles of Fe2+ are oxidized. Using the molarity and volume of FeSO4, the moles of Fe2+ can be calculated, which then allows for the calculation of the required moles of KMnO4. Finally, applying the molarity equation (M1V1 = M2V2) will yield the necessary volume of KMnO4 solution. Balancing the reaction is crucial for accurate calculations in redox titrations.
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1. What volume of a 0.0200 M KMnO4 solution is required to oxidize 40.0 mL of 0.100 M FeSO4 in sulfuric acid solution?



2. MnO4-1 + H+ + Fe+2 --> Fe+3 + Mn+2 + H2O



3. I honestly have no idea how to tackle this, I failed the titration portion of the AP Exam. I Think is maybe M1V1=M2V2 though

1. A 20.00 mL sample of Na2SO3 was titrated with 36.0 mL of 0.0513O M K2Cr2O7 solution in the presence of H2SO4. Calculate the molarity of the Na2SO3 solution.



2. SO3-2 + Cr2O7 + H+ --> SO4-2 + Cr+3 + H2O



3. No idea.

As a note, I didn't know how to do superscripts and subscripts, so the tiny ones are the amount of that element is in that cmpd, and the big italicized numbers are the charge of the ion or element.
 
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I would calculate how many moles of FeSO_{4} and from that calculate how many moles of KMNO_{4} would be needed to oxidize the compound.
 
Most important thing: you have to balance reaction equations. Without that you won't move ahead.

Superscripts and subscripts are done with [ sub][ /sub] and [ sup][ /sup] tags (no spaces after opening square bracket)
 
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