How to balance a chemical equation

In 2H₂O, the large 2 is a coefficient: it multiplies the whole formula. The term represents four hydrogen atoms and two oxygen atoms. The small 2 in H₂O is a subscript: each water molecule contains two hydrogen atoms. A missing coefficient or subscript means one.

Start from the formulas supplied in the question. For H₂ + O₂ → H₂O, the left has two hydrogen atoms and two oxygen atoms; the right has two hydrogen atoms and one oxygen atom. Changing H₂O to H₂O₂ would change water into a different substance, rather than balance the water equation.

Put a coefficient of 2 before H₂O to match oxygen. That gives four hydrogen atoms on the right, so put 2 before H₂ on the left. The result is 2H₂ + O₂ → 2H₂O. Count again: both sides now contain four hydrogen atoms and two oxygen atoms.

Use Al + O₂ → Al₂O₃ as a practice equation. An initial count gives Al: left 1, right 2; O: left 2, right 3. Oxygen occurs in pairs on the left and groups of three on the right. Six is the smallest positive number divisible by both two and three.

Put 3 before O₂ and 2 before Al₂O₃. The equation is now Al + 3O₂ → 2Al₂O₃. Oxygen balances at six atoms on each side, but the product now contains 2 × 2 = 4 aluminium atoms. Put 4 before Al.

The balanced equation is 4Al + 3O₂ → 2Al₂O₃. Your final count should read Al: left 4, right 4; O: left 6, right 6. Recount every element after changing a coefficient, because one coefficient can affect more than the element you were trying to fix.

When an element appears in several substances, add all its contributions. Consider Fe₂O₃ + CO → Fe + CO₂. Start by putting 2 before Fe, matching the two iron atoms in Fe₂O₃. Carbon is initially one on each side, but oxygen is four on the left and two on the right.

Try a coefficient of 3 before CO. To keep carbon balanced, put 3 before CO₂ too. The result is Fe₂O₃ + 3CO → 2Fe + 3CO₂. Oxygen on the left is 3 from Fe₂O₃ plus 3 from the three CO units, making 6. Oxygen on the right is 3 × 2 = 6. Carbon is 3 on each side, and iron is 2 on each side.

Bracketed groups need the same care. In Ca(OH)₂, the subscript outside the bracket applies to both O and H, so there are one calcium, two oxygen and two hydrogen atoms. In 3Ca(OH)₂ there are three calcium, six oxygen and six hydrogen atoms. Count the formula first, then multiply by the coefficient.

If every coefficient has a common factor, divide all of them by it. For example, 4H₂ + 2O₂ → 4H₂O has equal atom counts but reduces to 2H₂ + O₂ → 2H₂O. The coefficients express a ratio; the usual final form uses the smallest whole numbers.

Equal atom counts do not require equal numbers of molecules. In the water equation, two units of hydrogen and one of oxygen produce two units of water. The total number of molecules changes, while the numbers of hydrogen and oxygen atoms remain equal across the arrow.

For an ionic equation, check total charge as well as atoms. Equal atom counts alone cannot confirm an ionic equation is balanced. For equations involving electrons or more complicated redox changes, use the method required by your task; do not assume trial-and-error coefficients have checked charge for you.

Copy this working record: original equation; elements present; initial left and right counts; coefficients tried; final counts for every element; common factor removed. For Fe₂O₃ + 3CO → 2Fe + 3CO₂, the final lines are Fe 2/2, C 3/3 and O 6/6, where each pair means left/right.

Practise with Mg + O₂ → MgO before checking the answer. Oxygen starts at 2/1. Putting 2 before MgO makes it 2/2, then putting 2 before Mg balances magnesium. The result is 2Mg + O₂ → 2MgO, with Mg 2/2 and O 2/2.

If your counts disagree, inspect the formula that contributes to that element. Check whether you missed a coefficient, a bracket subscript or an additional substance on the same side. Keep the original formulas visible so an accidental subscript change does not pass unnoticed.