Answer:
41.9(w/w) %
Explanation:
Based on the reaction:
Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)
Where 1 mole of sodium oxalate reacts with 2 moles of HCl
Moles of HCl solution to reach end point are:
44.15mL = 0.04415L ₓ (0.250mol / L) = 0.01104 moles of HCl
As 2 moles of HCl reacts per mole of Na₂C₂O₄:
0.01104mol HCl ₓ (1 mol Na₂C₂O₄ / 2 mol HCl) = 5.519x10⁻³ moles Na₂C₂O₄ are in the sample.
Molar mass of Na₂C₂O₄ is 134g/mol; thus, mass of 5.519x10⁻³ moles Na₂C₂O₄ is:
5.519x10⁻³ moles Na₂C₂O₄ ₓ (134g / mol) = 0.740g of Na₂C₂O₄ in the sample.
Thus, percent by mass of sodium oxalate in the sample is:
0.740g of Na₂C₂O₄ / 1.766g ₓ 100 =
