Calculate the number of molecules of water in oxalic acid hydrate, from the following: 5.00 g of the coumpound is made up to 250 ml and 25 ml of the solution requires 15.90 ml of .5 M NaOH to reach neutralization;
What I have so far is the moles of NaOH: .00795
Moles of oxalic acid in 250 ml= .3702 mol and in 25 ml = .009255 mole.
If I subtract the amount of moles of acid (25 ml) from the amound of NaOH, I end up with .001305 mole of acid that is used up as it looks like a one to one ratio? no?
0 + NaOH -> H+
+ 2 CO2
Am I on the right track? but then what?
Thanks in advance