A chemical equation describes a reaction with formulae. The substances that react, the reactants, are written on the left, the substances that form, the products, on the right, and an arrow joins the two sides.
When hydrogen burns in oxygen, the molecules break apart and their atoms join again in new combinations. No atom is created or destroyed, so the equation must show the same number of each kind of atom on both sides: \mathrm{2H_2 + O_2 \longrightarrow 2H_2O}.
The number written in front of a formula is its coefficient. It counts whole molecules or formula units, so here 2 hydrogen molecules and 1 oxygen molecule give 2 water molecules. A coefficient of 1 is not written.
First write the correct formulae of all reactants and products, and never change them to make an equation balance. Changing \mathrm{H_2O} into \mathrm{H_2O_2} would describe hydrogen peroxide, a different substance, so only the coefficients may change.
For propane burning in air, give the most complicated formula, \mathrm{C_3H_8}, a coefficient of 1. Its 3 carbon atoms then need \mathrm{3CO_2} on the right, and its 8 hydrogen atoms need \mathrm{4H_2O}.
Oxygen is balanced last because it appears in two products. The right side now has 3 × 2 + 4 × 1 = 10 oxygen atoms, which needs \mathrm{5O_2}, giving \mathrm{C_3H_8 + 5O_2 \longrightarrow 3CO_2 + 4H_2O}.
Some equations are hard to balance by inspection. When copper dissolves in dilute nitric acid, copper(II) nitrate, nitrogen monoxide and water form, and oxygen appears in four of the substances.
Write the coefficients as letters, a\mathrm{Cu} + b\mathrm{HNO_3} \longrightarrow c\mathrm{Cu(NO_3)_2} + d\mathrm{NO} + e\mathrm{H_2O}, and write one equation for each element: a = c for copper, b = 2e for hydrogen, b = 2c + d for nitrogen and 3b = 6c + d + e for oxygen.
Putting a = 1 gives c = 1, and solving the other three equations gives b = 8/3, d = 2/3 and e = 4/3. Multiplying every coefficient by 3 removes the fractions: \mathrm{3Cu + 8HNO_3 \longrightarrow 3Cu(NO_3)_2 + 2NO + 4H_2O}.
When barium chloride solution is added to sodium sulfate solution, a white precipitate of barium sulfate forms: \mathrm{BaCl_2 + Na_2SO_4 \longrightarrow BaSO_4 + 2NaCl}.
In the solutions these compounds are present as ions. Only \mathrm{Ba^{2+}} and \mathrm{SO_4^{2-}} come together, while the sodium and chloride ions stay dissolved and unchanged. Ions that take no part in a reaction are called spectator ions.
An ionic equation leaves out the spectator ions and shows only the particles that change: \mathrm{Ba^{2+} + SO_4^{2-} \longrightarrow BaSO_4}. The same equation describes the reaction of any soluble barium salt with any soluble sulfate.
An ionic equation must balance electric charge as well as atoms. When a copper wire is placed in silver nitrate solution, crystals of silver grow on the wire and the solution slowly turns blue as copper(II) ions form.
Writing \mathrm{Cu + Ag^+ \longrightarrow Cu^{2+} + Ag} balances the atoms but not the charge, which totals 1+ on the left and 2+ on the right. Each copper atom gives up two electrons, but each silver ion takes only one.
Two silver ions are therefore needed for each copper atom: \mathrm{Cu + 2Ag^+ \longrightarrow Cu^{2+} + 2Ag}. Now the total charge is 2+ on both sides, and the copper and silver atoms still balance.