Problem help?

Question

In the back of my book we have extra problems we can do, to work prepart ourselves for the test. It gives answers, but no explanations. There were a few I didn't understand, so my chem teacher was supposed to help me during lunch. He was absent today (we have a block schedule, so we'll still have the test)! Anyway, I need help solving these equations. Any help would be great!

1) The concentration of a H2O2 solution can by determine by titration against a standardized potassium permanganate solution in an acidic medium according to the following equation:
2MnO4- + 5H2O2 + 6H+ --> 5O2 + 2Mn2+ + 8H2O
If 36.44mL of a 0.01625M KMnO4 solution are required to completely oxidize 25.00mL of a H2O2 solution, calculate the molarity of the H2O2 solution.

2) H2C2O4 is present in many plants and vegetables. If 24.0mL of 0.0100 M KMnO2 solution is needed to titrate 1.00g of a sample of H2C2O4 to the equivalence point, what is the percent mass of H2C2O4 in the sample?

the net ionic is:
2MnO4- + 16H+ + 5(C2O4)2- ---> 2Mn2+ + 10CO2 + 8H2O

Thanks for any help!

I really don't understand question #2, but thats what the book gives me.

Answer 1

The molarity of the H2O2 solution will be in units of molH2O2/1000mLH2O2. So begin with 36.44mL MnO4-/25.00mL H2O2. That comes from the titration data, and it contains mL H2O2 in the denominator that you will need later, because it appears in the answer. Next multiply by 0.01625molMnO4-/1000mL MnO4-. That comes from the molarity of the MnO4- solution. The mL MnO4- solution cancel, leaving molMnO4-. Next multiply by 5mol H2O2/2molMnO4-. That comes from the balanced equation. The molMnO4- cancel, leaving molH2O2/1000mL H2O2, which is the answer.

In part 2) of your question, you mention KMnO2, which does not exist, and you do not give a balanced equation. Please ask this question again correctly, and I shall try to help you. I am not allowed to answer a question that I have already answered.

Answer 2

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