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.

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.

@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?

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

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

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 = ?

To find the volume of 0.1 N KMnO4 solution required to titrate the given amount of K2[Cu(C2O4)2]·2H2O, you need to follow several steps.

Step 1: Determine the molar mass of K2[Cu(C2O4)2]·2H2O:
The molar mass of K2[Cu(C2O4)2]·2H2O can be calculated by summing the atomic masses of all the elements present in the compound:
K (2 atoms) + Cu (1 atom) + C (4 atoms) + O (8 atoms) + H (4 atoms) = Molar mass of K2[Cu(C2O4)2]·2H2O

Step 2: Calculate the moles of K2[Cu(C2O4)2]·2H2O:
Using the molar mass and given weight of K2[Cu(C2O4)2]·2H2O, you can calculate the moles of the compound by dividing the weight by the molar mass:
moles = mass of K2[Cu(C2O4)2]·2H2O / molar mass

Step 3: Determine the stoichiometry between KMnO4 and K2[Cu(C2O4)2]·2H2O:
Based on the balanced chemical equation, you know that KMnO4 takes 5 electrons while oxalate (C2O4) takes 2 electrons to form CO2 when oxidized. This information will help you establish the stoichiometry between KMnO4 and K2[Cu(C2O4)2]·2H2O.

Step 4: Calculate the moles of KMnO4 required:
By considering the stoichiometry from Step 3, you can determine the moles of KMnO4 required to oxidize the moles of K2[Cu(C2O4)2]·2H2O calculated in Step 2.

Step 5: Convert moles of KMnO4 to volume:
Since you have the normality of KMnO4 (0.1 N), which implies that 1 liter (1000 mL) of the solution contains 0.1 mole of KMnO4, you can calculate the volume (in mL) of 0.1 N KMnO4 solution needed to react with the calculated moles of KMnO4 from Step 4.

By following these steps, you should be able to determine the volume of 0.1 N KMnO4 solution needed to titrate the given amount of K2[Cu(C2O4)2]·2H2O. Remember to use accurate molar masses and pay attention to stoichiometry when performing the calculations.