Calculate the percentage of uranium in the sample. Thank you!!?

2 MnO4{-} + 16 H{+} + 5 Na2C2O4 →
10 CO2(g) + 2 Mn{2+}(aq) + 8 H2O(l) + 10 Na{+}

(0.342 g Na2C2O4) / (133.9985 g Na2C2O4/mol) x
(2 mol MnO4{-} / 5 mol Na2C2O4) / (50.3 mL) = 2.0296 × 10^-5 mol MnO4{-} / mL

UO{2+}→UO2{2+} = U{4+} → U{6+}

5 U{4+} + 2 MnO4{-} + 16 H{+} → 2 Mn{2+} + 5 U{6+} + 8 H2O

(32.0 mL) × (2.0296 × 10^-5 mol MnO4{-} / mL) × (5 mol U / 2 mol MnO4{-}) ×
(238.02891 g U/mol) / (4.60 g total) = 0.084018 = 8.40% U by mass

First titration: Titration of Na₂C₂O₄ against MnO₄⁻:
2MnO₄⁻(aq) + 16H⁺(aq) + 5C₂O₄²⁻(aq) → 2Mn²⁺(aq) + 8H₂O(ℓ) + 10CO₂(g)

Molar mass of Na₂C₂O₄ = (23.0×2 + 12.0×2 + 16.0×4) g/mol = 134.0 g/mol
Moles of Na₂C₂O₄ used = (0.342 g) / (134.0 g/mol) = 0.002552 mol
Moles of C₂O₄²⁻ reacted = 0.002552 mol
Moles of MnO₄⁻ reacted = (0.002552 mol) × (2/5) = 0.001021 mol

Second titration: Titration of uranium-containing material against MnO₄⁻:
2MnO₄⁻(aq) + 6H⁺(aq) + 5UO²⁺(aq) → 2Mn²⁺(aq) + 3H₂O(ℓ) + 5UO₂²⁺(g)

Moles of MnO₄⁻ reacted = (0.001021 mol) × (32.0/50.3) = 0.0006495 mol
Moles of UO²⁺ reacted = (0.0006495 mol) × (5/2) = 0.001624 mol
Moles of U in the sample = 0.001624 mol
Molar mass of U = 238.0 g/mol
Mass of U in the sample = (0.001624 mol) × (238 g/mol) = 0.387 g
Mass percent of U in the sample = (0.387/4.60) × 100% = 8.41%

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OR:

First titration:
(0.342 g Na₂C₂O₄) × (1 mol Na₂C₂O₄ / 134.0 g Na₂C₂O₄) × (2 mol MnO₄⁻/ 5 mol Na₂C₂O₄) / (50.3/1000 L MnO₄⁻ solution)
= 0.0203 M MnO₄⁻

Second titration:
(0.0203 mol MnO₄⁻ / 1000 mL MnO₄⁻ solution) × (32.0 mL MnO₄⁻ solution) × (5 mol U / 2 mol MnO₄⁻) × (238 g U / 1 mol U)
= 0.387 g U

(0.387/4.60) × 100% = 8.41% U in the sample