\(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)

\(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2}\)

The balanced half equations are

\(\begin{aligned} \left(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow\right.\left.\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\right) \times 2 \\ \left(\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow\right.\left.2 \mathrm{CO}_{2}+2 \mathrm{e}^{-}\right) \times 5 \end{aligned}\)

Equating number of electrons, we get \(2 \mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+}+10 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}\)

\(5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow 10 \mathrm{CO}_{2}+10 \mathrm{e}^{-}\)

On adding both the equations, we get

\(\begin{aligned} 2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{-}+16 \mathrm{H}^{+}  \longrightarrow 2 \mathrm{Mn}^{2+} \\ +2 \times 5 \mathrm{CO}_{2}+\frac{16}{2} \mathrm{H}_{2} \mathrm{O} \end{aligned}\)

Thus, \(x, y\) and \(z\) are \(2,5\) and \(16,\) respectively.