1. Identify the half-reactions in acidic medium

   The reaction is between permanganate and oxalate in acidic medium.

   • Reduction half-reaction: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$

   • Oxidation half-reaction: $C_2O_4^{2-} \to 2CO_2 + 2e^-$

2. Equalize electrons exchanged

   The least common multiple of electrons $= 10$.

   • Multiply the manganese half-reaction by $2$: $2MnO_4^- + 16H^+ + 10e^- \to 2Mn^{2+} + 8H_2O$

   • Multiply the oxalate half-reaction by $5$: $5C_2O_4^{2-} \to 10CO_2 + 10e^-$

3. Add the half-reactions

   $2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \to 2Mn^{2+} + 8H_2O + 10CO_2$

4. Compare with the given general form

   Given: $xMnO_4^- + yC_2O_4^{2-} + zH^+ \to xMn^{2+} + 2yCO_2 + \frac{z}{2}H_2O$

   From the balanced equation:

   • $x = 2$
   • $y = 5$
   • $z = 16$

   Check:

   • $2y = 2(5) = 10$ gives $10CO_2$ ✔️
   • $\dfrac{z}{2} = \dfrac{16}{2} = 8$ gives $8H_2O$ ✔️

5. Evaluate options

   • A: $(5,2,16)$ ❌
   • B: $(2,5,8)$ ❌
   • C: $(2,5,16)$ ✅
   • D: $(5,2,8)$ ❌

Therefore, the correct answer is Option C.