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πŸ§ͺ Chemical Equation Balancer

Enter any unbalanced chemical equation. Get instant balanced coefficients and reaction types for Class 10 & 12 Boards.

Popular Board Exam Presets:
Rusting (Fe + H2O) Water Synthesis Respiration Methane Combustion Aluminium Oxide Lead Nitrate Thermal Limewater Test Redox (KMnO4 + HCl)
Balanced Chemical Equation:
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Why equations must balance

Balancing rests on one law: the law of conservation of mass. Atoms are neither created nor destroyed in a chemical reaction β€” they only rearrange. So the count of each element must be identical on both sides. A balanced equation is not a style choice; it is the only chemically honest description of the reaction, and it is the first thing a board evaluator checks before awarding any marks to the answer.

The method this balancer uses: trial with inspection order

  1. Write correct formulas first. Hβ‚‚O stays Hβ‚‚O β€” never change subscripts to balance; only coefficients change.
  2. Count atoms per element on both sides. Polyatomic ions that appear unchanged on both sides (like SO₄²⁻ or NO₃⁻) can be balanced as one unit.
  3. Balance the most complex compound first, then elements appearing in single compounds, and leave free elements (Hβ‚‚, Oβ‚‚, metals) for last.
  4. Clear fractions by multiplying the whole equation, then recount everything once.

Worked example β€” burning propane

Unbalanced: C₃Hβ‚ˆ + Oβ‚‚ β†’ COβ‚‚ + Hβ‚‚O. Balance carbon: 3 COβ‚‚. Balance hydrogen: 4 Hβ‚‚O. Now count oxygen: right side has 3Γ—2 + 4Γ—1 = 10, so Oβ‚‚ gets coefficient 5. Final: C₃Hβ‚ˆ + 5Oβ‚‚ β†’ 3COβ‚‚ + 4Hβ‚‚O. Count again: 3 C, 8 H, 10 O on each side β€” balanced.

Notice the order: C and H (bound in single compounds) first, O (free molecule) last. That ordering solves most textbook equations in under a minute.

What balancing unlocks next

Balanced equations are the start of stoichiometry: the coefficients give mole ratios, which convert between mass, moles, volume of gases at STP (22.4 L per mole) and limiting-reagent questions. A typical board numerical β€” β€œwhat mass of COβ‚‚ forms when 11 g of propane burns?” β€” flows directly from those coefficients: 44 g propane β†’ 3 Γ— 44 = 132 g COβ‚‚, so 11 g gives 33 g.

Reaction types matter too: combination, decomposition, displacement, double displacement, and redox. The balancer's output notes the reaction type so you can file examples under the right heading before the exam.

Common mistakes to avoid

  • Changing subscripts (Hβ‚‚O β†’ Hβ‚‚Oβ‚‚) to force a balance β€” that changes the chemical itself.
  • Balancing O or H first in combustion equations β€” always the slow path.
  • Forgetting to recount after the final coefficient β€” one wrong count undoes the whole answer.

Stuck on a specific equation or a stoichiometry numerical? Send it as a photo on WhatsApp and get the steps explained at your class level.