5C2O42- + 2MnO4- + 16H+ ?10CO2 + 2Mn2+ + 8H2O A 0.250 g sample of Na2C2O4 requir

Question

5C2O42- + 2MnO4- + 16H+ ?10CO2 + 2Mn2+ + 8H2O

A 0.250 g sample of Na2C2O4 requires 22.15 mL of a certain KMnO4 solution to reach the end point during a titration.  What is the concentration of the KMnO4 solution in moles/liter?  Do not include units in your answer.

A 0.750 g sample of an unknown solid is dissolved in 100 mL of water and acidfied with 25 mL of 3 M H2SO4 then titrated with a 0.0200 M KMnO4 solution. If the unknown solid requires 12.5 mL of the KMnO4 solution to reach the endpoint, what is the % sodium oxalate in the unknown solid? Assume the only compound in the unknown solid that reacts with the KMnO4 solution is sodium oxalate. Report you answer as a percentage, but do not include the percent sign, %, in your answer

Explanation / Answer

The carbon in the C2O4 -2, oxalate, is being oxidized from an oxidation # of +3 for each carbon ... into +4 for each carbon

Manganese @ +7 each in the MnO4- is reduced, having its oxidation number reduced down to +2 each

What is the role of each of the reactants/products
MnO4- (the oxidizer, it takes electrons)
H+ (combines with MnO4 - 's oxygens to make water)
C2O42- (the reducer, it loses electrons)
Mn2+ (the reduced form of manganese formed by MnO4- taking electrons)
H2O (molecule formed with MnO4-'s oxygens)
CO2 (molecule formed with the oxidized form of carbon after oxalate lost electrons)

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