Chemical Equation Balancing — Rules and Worked Examples
An unbalanced equation is a wrong equation. Every stoichiometry numerical you will ever attempt — limiting reagent, percentage yield, volume of gas produced, titration — starts from a balanced equation. If the balancing is wrong, every step after it is wrong too, and in board exams the balanced equation itself usually carries a mark of its own. This guide gives you the rules, the order to apply them in, and four worked examples.
Why we balance: the law of conservation of mass
Atoms are not created or destroyed in a chemical reaction. They are only rearranged. So the number of atoms of each element on the left (reactants) must equal the number on the right (products).
We achieve this by putting whole numbers called coefficients in front of the formulas. The coefficient multiplies every atom in that formula. In 2H₂O the coefficient 2 gives 4 H atoms and 2 O atoms.
The six rules
- Never change a subscript. Changing H₂O to H₂O₂ turns water into hydrogen peroxide — a different substance. Only coefficients may change.
- Balance elements that appear in only one reactant and one product first. Usually that means carbon, then hydrogen.
- Leave free elements (O₂, H₂, Cl₂, Fe) for last. They can absorb any leftover imbalance without disturbing anything else.
- Treat an unchanged polyatomic ion as a single unit. If SO₄²⁻ appears intact on both sides, balance "sulfate" as one block, not S and O separately.
- Fractions are allowed while working — clear them at the end by multiplying the whole equation by the denominator.
- Reduce to the smallest whole numbers. 2C + 2O₂ → 2CO₂ is correct arithmetic but is not the accepted answer; C + O₂ → CO₂ is.
Worked example 1 — Combustion of methane
Skeleton: CH₄ + O₂ → CO₂ + H₂O
Carbon: 1 on the left, 1 on the right. Already balanced.
Hydrogen: 4 on the left, only 2 on the right. Put 2 before H₂O → 4 H on the right.
Oxygen last: right side now has 2 (in CO₂) + 2 (in 2H₂O) = 4 O atoms. Left side needs 4 O,
so put 2 before O₂.
CH₄ + 2O₂ → CO₂ + 2H₂O
Check — C: 1 = 1 · H: 4 = 4 · O: 4 = 4. ✔
Worked example 2 — Combustion of propane
Skeleton: C₃H₈ + O₂ → CO₂ + H₂O
Carbon: 3 on the left → put 3 before CO₂.
Hydrogen: 8 on the left → 8 H needs 4 water molecules → put 4 before H₂O.
Oxygen last: right side has 3 × 2 = 6 (from CO₂) plus 4 × 1 = 4 (from H₂O) = 10 O atoms.
Left side supplies O in pairs, so 10 ÷ 2 = 5 gives the coefficient of O₂.
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Check — C: 3 = 3 · H: 8 = 8 · O: 10 = 6 + 4 = 10. ✔
Worked example 3 — Aluminium with hydrochloric acid
Skeleton: Al + HCl → AlCl₃ + H₂
Chlorine appears as 1 on the left and 3 on the right, and hydrogen comes in 2s on the right. The smallest number divisible by both 3 and 2 is 6, so aim for 6 Cl and 6 H.
Put 6 before HCl → 6 H and 6 Cl on the left.
6 Cl on the right needs 2 AlCl₃ (2 × 3 = 6).
2 AlCl₃ needs 2 Al on the left.
6 H on the right needs 3 H₂ (3 × 2 = 6).
2Al + 6HCl → 2AlCl₃ + 3H₂
Check — Al: 2 = 2 · H: 6 = 6 · Cl: 6 = 6. ✔
Worked example 4 — The fraction trick (ethane)
Some equations refuse to balance in whole numbers on the first pass. Use a fraction, then clear it.
Skeleton: C₂H₆ + O₂ → CO₂ + H₂O
Carbon: 2 → put 2 before CO₂. Hydrogen: 6 → put 3 before H₂O.
Oxygen on the right = 2 × 2 + 3 × 1 = 7 atoms. But O₂ supplies even numbers only, so the
coefficient must be 7 ÷ 2 = 7/2.
C₂H₆ + (7/2)O₂ → 2CO₂ + 3H₂O
Multiply every coefficient by 2:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Check — C: 4 = 4 · H: 12 = 12 · O: 14 = 8 + 6 = 14. ✔
A quick word on redox equations
Reactions where oxidation states change — permanganate titrations, dichromate oxidations, disproportionation — are far faster to balance by the ion-electron (half-reaction) method than by inspection. You first assign oxidation numbers to find what is oxidised and what is reduced, then balance electrons lost against electrons gained. Inspection still works, but it is slow and error-prone in an exam. Learn oxidation numbers first; the balancing follows almost automatically.
Common mistakes that cost marks
- Changing subscripts to force a balance. Writing CH₄ + O₂ → CO₂ + H₄O₂ "balances" but describes substances that do not exist. Instant zero.
- Balancing oxygen first. Oxygen usually appears in three or four species at once, so fixing it early forces you to redo everything. Always last.
- Forgetting the coefficient multiplies the whole formula. In 3Ca(OH)₂ there are 3 Ca, 6 O and 6 H — not 3 Ca, 2 O and 2 H.
- Leaving fractions in the final answer. A fraction is a legal working step, never a legal final equation in a school or entrance exam.
- Not reducing. If every coefficient is divisible by a common factor, divide through.
- Skipping the final check. Ten seconds counting atoms element by element catches almost every slip.
Where balancing appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 9–10 | Writing and balancing equations for named reactions; combustion |
| CBSE/ICSE Class 11–12 | Stoichiometry, limiting reagent, redox by ion-electron method |
| JEE / NEET | Mole-ratio numericals where a wrong coefficient wrecks the whole answer |
| IIT-JAM / CUET-PG | Redox titration equivalents, permanganate and dichromate reactions |
| GATE / CSIR-NET | Balanced equations in inorganic mechanisms and quantitative analysis |
Check every equation in one click. The Chemical Equation Balancer takes your skeleton equation and returns the smallest whole-number coefficients, so you can confirm your own working instead of guessing.
Open the Chemical Equation Balancer →Want the reasoning taught properly rather than just the answer? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in the Delhi-NCR area — details at abcchemistry.in.