$$H_2C_2O_4 \cdot 2H_2O_{(aq)} + 2NaOH_{(aq)} \rightarrow Na_2C_2O_4_{(aq)} + 4H_2O_{(l)}$$

Moles of NaOH used:

$$Moles \ of \ NaOH = Concentration \times Volume$$

$$Moles \ of \ NaOH = 2.02 \ M \times 39.40 \ mL = 79.668\times10^{-3} \ mol$$

Moles of Oxalic Acid Hydrate:

$$Moles \ of \ H_2C_2O_4 \cdot 2H_2O = \frac{Mass}{Molar \ Mass}$$

$$Moles \ of \ H_2C_2O_4 \cdot 2H_2O = \frac{5.012 \ g}{126.07 \ g/mol} = 39.755\times10^{-3} \ mol$$

Ratio of Moles:

From the balanced chemical equation, we can see that 1 mole of oxalic acid hydrate reacts with 2 moles of NaOH. Therefore, the ratio of moles is 1:2.

$$\frac{Moles \ of \ H_2C_2O_4 \cdot 2H_2O}{Moles \ of \ NaOH} = \frac{39.755\times10^{-3} \ mol}{79.668\times10^{-3} \ mol} = 0.5$$

Since the ratio is not 1:2, it indicates that not all of the oxalic acid hydrate has reacted. There may be some other factors affecting the titration, such as the presence of impurities or incomplete reaction.