Analysis of Transition Metal Oxalate Complex Salt Hydrate?

(a)
RbₓCr(C₂O₄)₃ is a neutral compound. Total charges :
(+1)*x + (+3) + (-2)*3 = 0
x = 3
Hence, formula of the compounds is Rb₃Cr(C₂O₄)₃•wH₂O


The equation for the reaction between C₂O₄²⁻ and MnO₄⁻ (in an acidic medium):
2MnO₄⁻(aq) + 16H⁺(aq) + 5C₂O₄²⁻(aq) → 2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)
OR: Mole ratio MnO₄⁻ : C₂O₄²⁻ = 2 : 5

Number of moles of KMnO₄ solution = (0.02566 mol/L) × (18.70/1000 L) = 0.0004798 mol
Number of moles of MnO₄⁻ ions reacted = 0.0004798 mol
Number of moles of C₂O₄²⁻ ions reacted = (0.0004798 mol) × (5/2) = 0.001200 mol


(b)
Number of moles of C₂O₄²⁻ ions per g of the compound = (0.001200 mol) / (0.2505 g) = 0.004790 mol/g


(c)
1 mole of Rb₃Cr(C₂O₄)₃•wH₂O contains 3 moles of C₂O₄²⁻.
Number of moles of Rb₃Cr(C₂O₄)₃•wH₂O in the sample = (0.001200 mol) × (1/3) = 0.00040000 mol
Given: Mass of Rb₃Cr(C₂O₄)₃•wH₂O in the sample = 0.2505 g

Molar mass of Rb₃Cr(C₂O₄)₃•wH₂O = (0.2505 g) / (0.00040000 mol) = 626.3 g/mol


(d)
Molar mass of Rb₃Cr(C₂O₄)₃•wH₂O :
85.47×3 + 52.00 + (12.01×2 + 16.00×4)×3 + (1.01×2 + 16.00)w = 626.3
572.47 + 18.01w = 626.3
w = (626.3 - 572.47) / 18.01
w = 2.991 ≈ 3

Hence, w = 3

what is the balanced equation for the reaction between KMnO4 and C2O4^2-?

In order to find the moles oxalate from the titration you need to know this first