Consider the following redox reaction in an acidic medium: x MnO_4^...

The given redox reaction can be split into two half-reactions.

Oxidation half-reaction:
$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^{-}$

Reduction half-reaction:
$\text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

To balance the number of electrons, multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$:

$5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10e^{-}$
$2\text{MnO}_4^{-} + 16\text{H}^{+} + 10e^{-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

Adding the two balanced half-reactions gives the overall balanced equation:

$2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}$

Comparing this with the given equation $x \text{MnO}_4^{-} + y \text{C}_2\text{O}_4^{2-} + z \text{H}^{+} \rightarrow x \text{Mn}^{2+} + 2y \text{CO}_2 + \dfrac{z}{2} \text{H}_2\text{O}$, we get:
$x = 2$
$y = 5$
$z = 16$

Answer: $2, 5, 16$