To balance a chemical equation, change the numbers in front of the formulas until every element has the same atom count on both sides. Keep the formulas themselves unchanged. Then reduce the coefficients to the smallest positive whole-number ratio.
This balancing chemical equations practice set helps you work out what to change when an equation won’t settle. You’ll use an atom ledger, try eight problems, and check answers that explain the next move. The formulas and products are supplied, so you can concentrate on balancing rather than predicting reactions.
Check what each number actually counts
A coefficient multiplies the entire formula that follows it. A subscript belongs to the formula. In 3H2O, the coefficient 3 gives six hydrogen atoms and three oxygen atoms. Changing that coefficient changes the amount of water represented. Changing the subscript changes the substance.
For ordinary chemical reactions, atom counts must match element by element. Matching the total number of atoms without checking their identities isn’t enough. OpenStax’s explanation of chemical equations connects this requirement to conservation of matter and distinguishes coefficients from subscripts.
Try a quick counting check before the worksheet: how many atoms appear in 2Al(OH)3? There are two aluminum atoms, six oxygen atoms, and six hydrogen atoms. The 3 outside the parentheses applies to both O and H. The coefficient 2 then multiplies all those counts. If you got three oxygens, the difficulty is counting, so fix that before choosing coefficients.
A balancing method you can check on paper
- Copy every formula carefully. Leave a space before each one for its coefficient.
- List the elements down the left of your page. Add columns for reactant and product counts.
- Start with an element that occurs in only one formula on each side, when possible.
- Change one coefficient, then update every count affected by that change.
- Continue until every row matches. Divide all coefficients by a common factor if needed.
This is a useful starting routine, not a promise that one element must always come first. Victor Valley College’s balancing lesson recommends beginning with elements appearing in one formula per side and checking all atoms after adjustments.
When the same polyatomic ion stays intact on both sides, you can temporarily count that group as a unit. Still check the individual atoms at the end. If the group breaks apart, counting it as an unchanged unit won’t work.
Worked example: aluminum and chlorine
Start with Al + Cl2 → AlCl3. The chlorine counts are two and three. Their smallest shared multiple is six, so put 3 before Cl2 and 2 before AlCl3. That creates two aluminum atoms on the right, requiring 2Al on the left.
Balanced equation: 2Al + 3Cl2 → 2AlCl3. The final ledger has Al: 2 and 2; Cl: 6 and 6. The coefficients 2, 3, and 2 share no factor greater than one. Notice that changing AlCl3 to AlCl2 would be an incorrect shortcut, even if the counts looked easier.
Eight practice problems
Write a coefficient in every blank, including 1 while you’re working. Try each equation before reading the answer section. These are paper exercises, not instructions to perform reactions. Physical states and reaction conditions are omitted because the task is atom counting.
- ___ Mg + ___ O2 → ___ MgO
- ___ N2 + ___ H2 → ___ NH3
- ___ KClO3 → ___ KCl + ___ O2
- ___ C3H8 + ___ O2 → ___ CO2 + ___ H2O
- ___ Fe2O3 + ___ CO → ___ Fe + ___ CO2
- ___ Na3PO4 + ___ CaCl2 → ___ Ca3(PO4)2 + ___ NaCl
- ___ C2H6 + ___ O2 → ___ CO2 + ___ H2O
- ___ Al + ___ HCl → ___ AlCl3 + ___ H2
Hints if you’re stuck
For problems 1 through 3, look for an element whose counts need a shared multiple. For 4 and 7, balance carbon, then hydrogen, then oxygen. In 5, oxygen appears in two reactants, so you must add those contributions. For 6, keep phosphate together initially. In 8, matching chlorine first leaves a hydrogen count that suggests doubling the aluminum-containing formulas.
Worked answers and atom checks
1. Magnesium oxide
2Mg + O2 → 2MgO. Oxygen arrives in pairs on the left, so the product needs a coefficient of 2. That change also creates two magnesium atoms on the right. Put 2 before Mg to match them. Final counts: Mg 2 on each side, O 2 on each side. Leaving Mg at 1 is a common unfinished answer.
2. Ammonia
N2 + 3H2 → 2NH3. Two nitrogen atoms require two ammonia molecules. Those contain six hydrogen atoms, supplied by three H2 molecules. Final counts: N 2, H 6 on each side. The coefficient 2 before ammonia multiplies its hydrogen subscript too; it doesn’t apply only to nitrogen.
3. Potassium chlorate decomposition
2KClO3 → 2KCl + 3O2. Oxygen occurs in groups of three and two, so aim for six oxygen atoms. That requires two chlorate formula units and three oxygen molecules. Two KCl units then balance potassium and chlorine. Final counts: K 2, Cl 2, O 6. If you wrote coefficients 4, 4, and 6, your counts match, but divide all three by 2.
4. Propane combustion
C3H8 + 5O2 → 3CO2 + 4H2O. Three carbons call for 3CO2; eight hydrogens call for 4H2O. The products now contain 6 + 4 = 10 oxygen atoms. Five O2 molecules supply them. Final counts: C 3, H 8, O 10. Counting only the oxygen in carbon dioxide would give an incomplete ledger.
5. Iron oxide and carbon monoxide
Fe2O3 + 3CO → 2Fe + 3CO2. Set iron to 2 on the right. The CO and CO2 coefficients must match because each contains one carbon. If that shared coefficient is x, the oxygen equation is 3 + x = 2x, giving x = 3. Check oxygen: the left has 3 + 3 = 6; the right has 3 times 2 = 6. This small algebra step is useful when inspection starts looping.
6. A phosphate group that stays together
2Na3PO4 + 3CaCl2 → Ca3(PO4)2 + 6NaCl. One product unit needs two phosphates and three calcium atoms. That fixes the first two coefficients at 2 and 3. The left then has six sodium atoms and six chlorine atoms, requiring 6NaCl. Final counts: Na 6, P 2, O 8, Ca 3, Cl 6. The parentheses in the product double the entire phosphate group.
7. Ethane combustion with a temporary fraction
2C2H6 + 7O2 → 4CO2 + 6H2O. Starting with one ethane gives 2CO2 and 3H2O. That’s seven oxygen atoms, or 7/2 O2. Multiply every coefficient by 2 to clear the fraction. Final counts: C 4, H 12, O 14. Doubling only the oxygen coefficient breaks the equation.
8. Aluminum and hydrochloric acid
2Al + 6HCl → 2AlCl3 + 3H2. Two aluminum chloride units contain six chlorines, supplied by 6HCl. Those six hydrogens form 3H2. Balance aluminum with 2Al. Final counts: Al 2, H 6, Cl 6. A coefficient of 6 before hydrogen gas would represent twelve hydrogen atoms, not six.
Turn a wrong answer into a useful study note
Keep three short lines for each miss: the equation you wrote, the first atom row that failed, and the coefficient change that repaired it. For example: “Problem 4: I counted oxygen only in CO2. Add the four oxygens in water before choosing O2.” That’s a more useful reminder than “be careful.”
On your next review, cover the answers and redo the missed equations on a blank page. Explain one coefficient choice aloud before checking. Use the chemistry exam study guide to fit this practice into your wider course review.
Compare notation with Chemistry notes, or browse study documents for related class materials. Check the course level and the author’s working before treating any shared answer as correct.
Once you’ve checked your ledger, upload your own notes to help other students. Include your explanations and corrections, and share only materials you’re allowed to distribute. A page showing why an answer works can be more helpful than a long answer list.
Frequently asked questions
Can I change a subscript to balance an equation?
No. A subscript is part of a substance’s formula. Change coefficients, then recount the atoms. If a supplied formula seems wrong, check the original question rather than silently rewriting it.
Are fractional coefficients always wrong?
A fraction can be a useful intermediate step. For these exercises, finish with the smallest whole-number coefficients. Multiply the entire equation by the denominator, then check whether a common factor remains.
Does a balanced equation prove a reaction will occur?
No. Balancing checks the accounting for the supplied reactants and products. It doesn’t establish reaction feasibility, rate, conditions, or product identity. Those are separate chemistry questions.
What if every count matches but my answer differs from the key?
Compare ratios. Coefficients 4, 2, and 4 describe the same ratio as 2, 1, and 2, but the latter is reduced. Also check that both answers use exactly the same formulas.