Consider the following reaction: xMnO₄⁻ + yC₂O₄²⁻ + zH⁺ → xMn²⁺ + 2yCO₂ + z/2H₂O. The values of x, y, and z in the reaction are, respectively:

On balancing the charge on both sides of reaction, we get
2x - 2y - z = 0
On balancing the oxidation number of Mn, we get
x(7-2) = 2x = 5x-2x = 3x
On balancing the oxidation number of C, we get
2y(4-2) = 4y
On balancing the oxidation number of O, we get
4x + 4y = 2y + 1/2 z
On balancing the number of atoms, we get x = x, 4x + 2y = 2y + z/2, and z = z.
Solving the equations, we get x = 2, y = 5, z = 16.