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The reaction between sodium hydroxide (NaOH) and oxalic acid (H2C2O4) is as follows :
$ \text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} $
We can see that 1 mole of oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) reacts with 2 moles of sodium hydroxide (NaOH).
Given that 50 mL of 0.5 M oxalic acid is needed to neutralize the sodium hydroxide, we can find the number of moles of oxalic acid that reacted :
$ \text{Moles of H}_2\text{C}_2\text{O}_4 = \text{Molarity} \times \text{Volume (L)} = 0.5 \, \text{mol/L} \times 0.05 \, \text{L} = 0.025 \, \text{mol} $
Since 1 mole of $\text{H}_2\text{C}_2\text{O}_4$ reacts with 2 moles of $\text{NaOH}$, the number of moles of $\text{NaOH}$ in 25 mL solution would be twice that of $\text{H}_2\text{C}_2\text{O}_4$ :
$ \text{Moles of NaOH} = 0.025 \, \text{mol} \times 2 = 0.05 \, \text{mol} $
Now, to find the mass of NaOH, we multiply the number of moles by the molar mass of NaOH (40 g/mol) :
$ \text{Mass of NaOH} = \text{Number of moles} \times \text{Molar mass} = 0.05 \, \text{mol} \times 40 \, \text{g/mol} = 2 \, \text{g} $
However, the question asks for the amount of $\text{NaOH}$ in 50 mL of the given solution. Since we've found the amount in 25 mL, we just need to double our result to find the amount in 50 mL :
$ 2 \, \text{g} \times 2 = 4 \, \text{g} $
So, the correct answer is Option B : 4 g.
