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Topic: Second Marathon Problem (Ridiculously Hard)  (Read 10393 times)

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Offline narutoverse13

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Re: Second Marathon Problem (Ridiculously Hard)
« Reply #15 on: September 20, 2014, 02:54:56 PM »
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.
« Last Edit: September 20, 2014, 03:05:06 PM by narutoverse13 »

Offline Borek

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Re: Second Marathon Problem (Ridiculously Hard)
« Reply #16 on: September 20, 2014, 03:10:14 PM »
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).
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Offline narutoverse13

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Re: Second Marathon Problem (Ridiculously Hard)
« Reply #17 on: September 20, 2014, 03:12:31 PM »
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!

Offline KungKemi

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Re: Second Marathon Problem (Ridiculously Hard)
« Reply #18 on: March 11, 2017, 06:44:00 PM »
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) + 5C2O42-(aq) ::equil:: 2Mn2+(aq) + 8H2O(l) + 10CO2 (g)

2MnO4-(aq) + OH-(aq) + 3CHO2-(aq) ::equil:: 3CO32-(aq) + 2H2O(l) + 2MnO2 (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

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