What volume of .1 N KMnO4 would be required to titrate .56g of K2[Cu(C2O4)2] 2H20? So, my problem is the normality thing and I need a general direction to go in if possible. From my research, the .1 N means that 1/10 of a mole of KMnO4 is dissolved. I know that KMnO4 takes 5 electrons and that oxalate takes 2 electrons to get CO2 when oxidized.

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  1. One way to do this is to convert 0.1N KMnO4 to M. To do that 0.1/5 = 0.02 M and work the problem as any stoichiometry problem.By the way your research is not right. 0.1 N means 1/10 equivalent weights are dissolved in 1L solution. An equivalent for KMnO4 is 158/5 = 31.6 grams but that;s just a close approximation.

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  2. @DrBob222 (.56g)(1mol K2[Cu(C2O4)2] 2H20/353.812g)(2 mol C2O4/1 mol K2[Cu(C2O4)2] 2H2O)(2mol KMnO4/5mol C2O4)(1L/.02mol KMnO4)= .063L

    Does this seem kosher?

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  3. I worked the problem and got 63.31 mL. I used 353.81 for the molar mass of the Cu compound. I worked it with normality and with molarity and obtained the same answer. In my opinion it really is a shame that no one teaches normality anymore

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  4. Yeah we only really touched on it for a lab where we had to find eq. wts of an unknown acid by titrating with a standardized NaOH solution

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  5. For whatever it's worth, when i was in undergraduate school and took quant we never even mentioned molar after the first week or so. Everything was normal. Using normality makes it so easy to calculate. It's
    mL x N x milliequivalent weight = grams. The only real question here, since all of the others are given in the problem, is "how do we calculate the milliequivalent weight". In this case it is 1/4 x molar mass = ?

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