Chemistry and laboratory
Chemical equation balancer
Balance chemical and ionic equations with exact whole-number coefficients, an atom-by-atom check and clear reasons when an equation cannot balance.
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Result
- Balanced equation
- 2 KMnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 8 H₂O + 5 Cl₂
- Coefficients
- 2 : 16 : 2 : 2 : 8 : 5
| Element | Reactant side | Product side |
|---|---|---|
| K | 2 | 2 |
| Mn | 2 | 2 |
| O | 8 | 8 |
| H | 16 | 16 |
| Cl | 16 | 16 |
- Coefficients are the smallest whole numbers. The balancer checks conservation only: it cannot tell whether the reaction actually happens.
Method and limits
Enter the reactants and products; the coefficients appear as you type. Redox and ionic equations work too: charge is balanced along with the atoms, and electrons can be written as e-.
Method: the null space of the composition matrix
Each species becomes a column and each element a row (plus a row for charge when ions are present). Entries are atom counts, positive for reactants and negative for products. Balancing means finding whole numbers x with A·x = 0. Gaussian elimination in exact fractions (no rounding) finds every solution:
- One independent solution with all coefficients positive: that is the balanced equation, scaled to the smallest whole numbers.
- No solution: some element (or the charge) cannot be conserved, usually because of a typo or a missing species. If an element appears on one side only, the tool names it.
- More than one independent solution: the equation combines separate reactions (for example H2 + O2 = H2O2 + H2O), so any mixture of them balances and there is no single answer.
- A solution that needs a zero or negative coefficient: a species cannot take part, or belongs on the other side.
A·x = 0, x = smallest positive integer vector of the one-dimensional null space
Limits
Balancing checks conservation only. It cannot say whether a reaction happens, what its products are, or whether the species exist. Write charges with ^ for anything other than ±1: Fe2+ is read as Fe₂ with a charge of one, so write Fe^2+.
Worked example
KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2 balances as 2 KMnO₄ + 16 HCl → 2 KCl + 2 MnCl₂ + 8 H₂O + 5 Cl₂. The ionic form MnO4^- + Fe^2+ + H^+ = Mn^2+ + Fe^3+ + H2O gives MnO₄⁻ + 5 Fe²⁺ + 8 H⁺ → Mn²⁺ + 5 Fe³⁺ + 4 H₂O, with a charge of +17 on each side.
References
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