Worked example 1: Balance propane combustion.

Start with C₃H₈ + O₂ → CO₂ + H₂O.
The formulas are already chemically correct, so only coefficients may change.
Put 3 in front of CO₂ and 4 in front of H₂O.
That matches the 3 carbon atoms and 8 hydrogen atoms coming from propane.
The products now contain 3 × 2 + 4 × 1 = 10 oxygen atoms, so use 5 O₂.
Oxygen is balanced last because it appears in more than one product.
Final equation: C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O.
Each element count now matches exactly.

Worked example 2: Balance the precipitation reaction Pb(NO₃)₂ + KI → PbI₂ + KNO₃.

Lead is already 1 to 1, so focus on iodide next.
PbI₂ needs two iodides on the product side.
Place 2 in front of KI.
That supplies the two iodides needed to make PbI₂.
Now place 2 in front of KNO₃.
Two potassium ions and two nitrate groups leave together in the soluble product.
Final equation: Pb(NO₃)₂ + 2 KI → PbI₂ + 2 KNO₃.
Treating nitrate as a repeating unit keeps the bookkeeping shorter and cleaner.

Worked example 3: Check charge in the simple redox net ionic line Zn + 2 H⁺ → Zn²⁺ + H₂.

Atom count: one Zn and two H atoms appear on both sides.
The element bookkeeping is satisfied.
Net charge left: +2. Net charge right: +2.
Ionic equations must conserve charge as well as atoms.
Check: a line with matching atoms but mismatched charge is still not balanced.
Charge bookkeeping is the extra rule molecular equations often hide.

Common mistake: balancing a net ionic equation by atom count alone and forgetting that the total charge on the left and right must also agree.

Chemical reactions