Consider the following reaction:
$$x \ MnO_4^- + y \ C_2O_4^{2-} + zH^+ \rightarrow x \ Mn^{2+} + 2y \ CO_2 + \frac{z}{2} H_2O$$
The values of x, y and z in the reaction are, respectively:

We first identify that the reaction takes place in an acidic medium, so we shall balance it by the ion-electron (half-reaction) method.

We have two half-reactions:

1. Reduction of permanganate:
$$MnO_4^- \rightarrow Mn^{2+}$$

2. Oxidation of oxalate:
$$C_2O_4^{2-} \rightarrow CO_2$$

Balancing the reduction half-reaction.

• Balance manganese atoms: already 1 on each side.
• Balance oxygen atoms by adding water. There are 4 oxygen atoms on the left, so we add 4 $$H_2O$$ to the right:
$$MnO_4^- \rightarrow Mn^{2+} + 4H_2O$$

• Balance hydrogen atoms by adding $$H^+$$. We now have 8 hydrogens on the right, so we add 8 $$H^+$$ on the left:
$$MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O$$

• Balance charge by adding electrons. The total charge on the left is $$(-1)+(+8)=+7$$. The total charge on the right is $$+2$$. To reduce the left side from +7 to +2 we add 5 electrons to the left:
$$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$

Balancing the oxidation half-reaction.

• Oxalate already has two carbon atoms, so we write two molecules of carbon dioxide on the right to keep carbon balanced:
$$C_2O_4^{2-} \rightarrow 2CO_2$$

• Oxygen is now balanced (4 on each side). No hydrogen appears, so no $$H^+$$ or $$H_2O$$ are needed here.

• Balance charge by adding electrons. The left side has a charge of $$-2$$. The right side is neutral. To make charges equal we add 2 electrons on the right:
$$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$$

Equalising the number of electrons. The reduction half needs 5 electrons, the oxidation half releases 2 electrons. Their least common multiple is 10, so we multiply:

• The reduction half-reaction by 2:
$$2MnO_4^- + 16H^+ + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$$

• The oxidation half-reaction by 5:
$$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$$

Adding the two half-reactions. Because both sides now contain 10 electrons, they cancel out when we add the equations:

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

Re-ordering the terms gives the balanced overall equation:

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

Comparing with the general form given in the question,

$$x \ MnO_4^- + y \ C_2O_4^{2-} + zH^+ \rightarrow x \ Mn^{2+} + 2y \ CO_2 + \frac{z}{2} H_2O,$$

we read directly:
$$x = 2,\; y = 5,\; z = 16.$$

Hence, the correct answer is Option A.