When balancing redox reactions under acidic conditions in aqueous solution, oxygen is balanced by adding:

When balancing redox reactions under acidic conditions in aqueous solution, oxygen is balanced by adding water as much as oxygen we have, on the opposite side, where the oxygens are.

Explanation:

We can see this redox balance, in order to determine what we said.

The unbalanced reaction is: MnO₂  +  HCl  →  MnCl₂ + Cl₂ + H₂O

The HCl determines, we are in acidic conditions.

We need to determine the half reactions:

Mn⁴⁺  →  Mn²⁺  (In this case, the oxidation number decreased)

This is the reduction. Mn caught 2 electrons

Mn⁴⁺  + 2e⁻  →  Mn²⁺

Cl⁻ → Cl₂

In ground state, the oxidation number is 0. The chloride was oxidized to chlorine. As Cl is a dyatomic molecule, we need 2 chlorides to balance so finally it released 2 electrons. This is the oxidation (ox. number has increased) 2Cl⁻ → Cl₂ + 2e⁻

Mn⁴⁺  + 2e⁻  →  Mn²⁺

2Cl⁻ → Cl₂ + 2e⁻

Mn⁴⁺  + 2Cl⁻ →  Mn²⁺ + Cl₂

Now we balance the oxygen. In reactant side, we have 2, so we complete with 2 water in the product side. We add water as much as oxygen we have, on the opposite side, where the oxygens are. (If we have 2 oxygen in reactant side, and we have 1 water, we need 1 more water)

So, as we added 2 H₂O, we complete with protons, on the opposite side

Mn⁴⁺  + 4H⁺  + 4Cl⁻ →  Mn²⁺ + 2Cl⁻ + Cl₂ + 2H₂O

(Notice, we had 2Cl⁻, but globally we have 4H⁺. At least, we have 4HCl)

The balanced equation is: MnO₂  + 4HCl →  MnCl₂ + Cl₂ + 2H₂O

the answer is 75.0%.