Chemistry Question?

Method 1 :

Molar mass of Na₂C₂O₄ = (23.0×2 + 12.0×2 + 16.0×4) g/mol = 134.0 g/mol

Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)
Mole ratio Na₂C₂O₄ : HCl = 1 : 2

Moles of HCl = (0.250 mol/L) × (44.15/1000 L) = 0.011038 mol
Mole of Na₂C₂O₄ = (0.011038 mol) × (1/2) = 0.005519 mol
Mass of Na₂C₂O₄ reacted = (0.005519 mol) × (134.0 g/mol) = 0.7395 g

% by mass of Na₂C₂O₄ in the sample = (0.7395/1.766) × 100% = 41.9%
(to 3 sig. fig. as 0.250 M HCl has 3 sig. fig.)

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Method 2:

(0.250 mol HCl / L) × (44.15/1000 L) × (1 mol Na₂C₂O₄ / 2 mol HCl) × (134.0 g Na₂C₂O₄ / 1 mol Na₂C₂O₄)
= 0.7395 g

(0.7395 g / 1.766 g) × 100% = 41.9% Na₂C₂O₄ in the sample
(to 3 sig. fig. as 0.250 M HCl has 3 sig. fig.)

Let me know if you need any clarification. I'd be glad to explain things more.

EDIT: I'm not sure if my jpeg shows up in my answer...let me know if you can't see it.