What is the balanced molecular equation of oxalic acid dihydrate and sodium hydroxide?

The dihydrate part of the reactant does not take part in the reaction. Throughout my training I was never taught to include it in any balanced equation. I do not think that students would be taught to do so today:
the balanced equation is:
(COOH)2 + 2NaOH → (COONa)2 + 2H2O

If you insist on writing oxalic acid as H2C2O4:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

And if you insist on including the dihydrate:
H2C2O4.2H2O + 2NaOH → Na2C2O4 + 4H2O

The problem with this last equation is that the reaction is carried out in aqueous medium, and the water of crystalisation or hydration of the oxalic acid is no longer associated with the oxalic acid when it is in solution. secondly: The product sodium oxalate occurs as a trihydrate, so if you have to consider the oxalic acid as a dihydrate, then in the equation you should consider that the sodium oxalate is a trihydrate, and you would get:
H2C2O4.2H2O + 2NaOH → Na2C2O4.3H2O + H2O - which I think is actually quite ridiculous.

My best advice is to stick with the very first equation that I submiited and you should be OK - or the second equation if your teacher uses that format for oxalic acid.

Oxalic Acid And Sodium Hydroxide

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RE:
What is the balanced molecular equation of oxalic acid dihydrate and sodium hydroxide?
H2C2O3*2H2O + NaOH --> ?

I keep trying it but i don't know if it makes sense. I know the first part is NaH2C2O4 but then do i add 4H2O?