[narutoverse13]

I actually got 11.6 mL of permanganate.
I got the concentration of permanganate to be .15 M. I did this calculations by finding moles of permanganate first and then dividing that by 18.55 mL.

[Borek]

Don't round down intermediate values (or rather: report them rounded, but use full precision or at least several guard digits in further calculations).

11.76 mL is what I got (note I am using EBAS, so my molar masses can be slightly different).

[narutoverse13]

Oh wow! I can't believe that I got so close to what you got!!!!! I am so excited!!!!!!!!! Thank you so much for your help Borek! You are right when you said that I may have rounded excessively. I couldn't have done this without your assistance!

[KungKemi]

Sorry to resurrect an old forum post, however, just for future reference, I did this question a few days ago and I found that for the second and third reactions you needed to balance these using the half-reaction method (which doesn't make a whole lot of sense because the Zumdahl textbooks only cover this for Electrochemistry later on). Anyway, in doing so you would get the equations:

2MnO₄^-_(aq) + 16H⁺_(aq) + 5C₂O₄^2-_(aq) 2Mn²⁺_(aq) + 8H₂O_(l) + 10CO_{2 (g)}

2MnO₄^-_(aq) + OH^-_(aq) + 3CHO₂^-_(aq) 3CO₃^2-_(aq) + 2H₂O_(l) + 2MnO_{2 (s)}

The first reaction occurs in acidic solution (because of the sulfuric acid), and the second reaction occurs in basic solution (because of the hydrolysis of water by the formate ion).

KungKemi