chemistry

Considerthe reaction between oxalate ions and acidified KMnO4 solution:
2MnO4^-(aq) + 16H^+(aq) + 5C2O4^2-(aq) → 2Mn^2+(aq)+ 8H2O(l) + 10CO2(g)
No. of moles of MnO4^- ions used = 0.00100 x (4.94/1000) = 0.00000494 mol
No. of moles of C2O4^2- ions = 0.00000494 x (5/2) =0.00001235 mol

Consider the precipitation of CaC2­O4:
Ca^2+(aq) + C2O4^2-(aq) → CaC2O4(s)
No. of moles of C2O4^- ions = 0.00001235 mol
No. of moles of Ca^2+ ions = 0.00001235 mol
Concentration of calcium in the serum = 0.00001235 / (5/1000) = 0.00247 mol/L

In the very first step, you must find out what reaction it is between KMnO4 and CaC2O4. We came across (MnO4)- as an oxidizing agent in HKCEE syllabus. So, would it be a redox reaction?

Let's suppose it is a redox. Try to balance the two half equations and then combine the two half equations, you must be able to find the mole ratio.

From HKCEE syllabus, we also learnt about molarity. We know that volume x conc = quantity.

With this and the mole ratio between (MnO4)- and (C2O4)2-, you can find the answer.