51.45 mL
We are given;
We are required to determine the volume of NaOH required.
The balanced equation for the reaction between NaOH and H₂C₂O₄ will be;
2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
We know that, moles are calculated by dividing the mass of the compound by its molar mass.
Moles = Mass ÷ Molar mass
Therefore;
Moles of H₂C₂O₄ = 3.587 g ÷ 90.0 g/mol
= 0.040 moles
From the equation, the mole ratio of NaOH to H₂C₂O₄ is 2 : 1
Therefore; Moles of NaOH = Moles of H₂C₂O₄ × 2
Thus;
Moles of NaOH = 0.040 moles × 2
= 0.080 moles
Molarity of NaOH = 1.555 M
But; Molarity = Moles ÷ Volume
Rearranging the formula;
Volume = Moles ÷ Molarity
Therefore;
Volume of NaOH = 0.080 moles ÷ 1.555 M
= 0.05145 L
But, 1 L = 1000 mL
Therefore;
Volume = 51.45 mL
Hence, the volume of HCl required is 51.45 mL