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Answer : 13.64 mL
Explanation : The reaction requires 13.64 mL of NaOH
From the reaction we know that oxalic acid requires two moles of NaOH so we have to calculate the number of moles of both.
0.150 L X 0.1500 moles/L = 2.25 X
So the moles of NaOH will be 2 X (2.25 X ) = 4.5 X
Now, we know the concentration of NaOH as 0.330 M
4.5 X moles / 0.330 moles/L = 0.01364 L or 13.64 mL
So, the volume of NaOH needed to neutralize oxalic acid will be 13.64 mL