UNIT 2 CHEMISTRY · PEARSON AAQ · C1.7
Balanced Equations for Period 3 Reactions
FREE CHEMISTRY REVISION · TOPIC 29
Build correct formulas, then conserve every atom.
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · C1.7
What you will learn
Construct balanced equations for the reactions in C1.3, C1.5 and C1.6. Start with chemically correct reactants and products, conserve each element and total charge, and add state symbols for the actual conditions. A balanced equation can still describe the wrong chemistry, so product selection comes first.
1. A reliable method
- Identify the reaction and conditions. Is it reaction with oxygen, water or chlorine, acid–base chemistry, dissolution or hydrolysis? Is water limited or in excess?
- Write correct formulas. Use ion charges or known molecular formulas. Keep Na₂O, Mg(OH)₂, Al₂Cl₆ and P₄O₁₀ intact.
- Count every atom on both sides. Multiply all subscripts inside brackets by the outside subscript, then multiply the complete formula by its coefficient.
- Change coefficients only. Balance the less common element first; leave H and O until later when convenient. An unchanged polyatomic group may be counted as a unit.
- Check the final equation. Count all atoms again, check total charge for ionic equations and reduce to the smallest whole-number coefficients.
- Add states and conditions. Use (s), (l), (g) and (aq) for solid, liquid, gas and aqueous species. An aqueous solution is not the same as a pure liquid.
A coefficient multiplies an entire chemical formula. 2Al(OH)₃ contains 2 Al atoms, 6 O atoms and 6 H atoms. Changing Al(OH)₃ to Al(OH)₂ changes the substance and cannot be used to balance an equation.
2. Start with correct formulas and states
| Species | Why the formula is correct | Common error |
|---|---|---|
| Na₂O | 2Na⁺ balance O²⁻ | NaO |
| Mg(OH)₂ | Mg²⁺ requires two OH⁻ ions | MgOH₂, which obscures the hydroxide grouping |
| Al₂O₃ | 2Al³⁺ balance 3O²⁻ | AlO or Al₃O₂ |
| Al₂Cl₆ | Specified molecular dimer representation | Losing the factor of 2 for Al or 6 for Cl |
| P₄O₁₀ | Specified phosphorus(V) oxide molecular formula | Replacing it with P₂O₅ when P₄O₁₀ is requested |
| Na₃PO₄ | 3Na⁺ balance PO₄³⁻ | NaPO₄ |
Hydrogen, oxygen and chlorine are H₂, O₂ and Cl₂ in these elemental reactions. Silicon is written Si(s), not Si₄. White phosphorus is represented by P₄(s); sulfur may be represented by S(s), or S₈(s) if explicitly specified. NaOH(aq) denotes dissolved sodium hydroxide, whereas Mg(OH)₂(s) denotes a sparingly soluble solid product.
3. C1.3 reactions: oxygen, water and chlorine
| Reaction set | Balanced equation | Condition or qualification |
|---|---|---|
| Oxygen | 4Na(s) + O₂(g) → 2Na₂O(s) | Specified oxide product; actual sodium combustion can form other oxygen compounds |
| Oxygen | 2Mg(s) + O₂(g) → 2MgO(s) | Burning magnesium |
| Oxygen | 4Al(s) + 3O₂(g) → 2Al₂O₃(s) | Protective oxide limits bulk reaction; finely divided metal reacts on heating |
| Oxygen | P₄(s) + 5O₂(g) → P₄O₁₀(s) | Sufficient oxygen; white phosphorus representation |
| Oxygen | S(s) + O₂(g) → SO₂(g) | If S₈ is specified: S₈ + 8O₂ → 8SO₂ |
| Water | 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g) | Cold water; vigorous |
| Water | Mg(s) + 2H₂O(l) → Mg(OH)₂(s) + H₂(g) | Slow with cold water; steam instead produces MgO |
| Water | 2Al(s) + 6H₂O(l) → 2Al(OH)₃(s) + 3H₂(g) | Formal hydroxide-producing equation when the protective layer is overcome; ordinary aluminium is passivated |
| Chlorine | 2Na(s) + Cl₂(g) → 2NaCl(s) | Chlorine supplies two Cl atoms per molecule |
| Chlorine | Mg(s) + Cl₂(g) → MgCl₂(s) | Metal-to-chloride reaction |
| Chlorine | 2Al(s) + 3Cl₂(g) → Al₂Cl₆ | Specified dimer product; phase depends on temperature |
| Chlorine | Si(s) + 2Cl₂(g) → SiCl₄(l) | Product state shown after cooling to room temperature |
| Chlorine | P₄(s) + 10Cl₂(g) → 4PCl₅(s) | Excess chlorine; product after cooling |
Equations are chemical bookkeeping, not a promise that every reaction is rapid under ordinary conditions. For aluminium and water, state the passivation qualification. A reaction conducted at high temperature may initially produce vapour even when a table shows the collected room-temperature product.
4. Worked example 1 — magnesium combustion
Task: Balance Mg + O₂ → MgO. The left side has two oxygen atoms; one MgO has only one. Put 2 before MgO, then balance the magnesium by putting 2 before Mg.
2Mg(s) + O₂(g) → 2MgO(s)
| Element | Left | Right |
|---|---|---|
| Mg | 2 | 2 |
| O | 2 | 2 |
The coefficient 1 before O₂ is understood. Do not change MgO to MgO₂: that would specify a different compound.
5. Worked example 2 — aluminium and water
Task: Balance the hydroxide-producing reaction Al + H₂O → Al(OH)₃ + H₂. First use 2Al and 2Al(OH)₃ to avoid a fractional hydrogen coefficient. The two hydroxide formulas need six oxygen atoms, so use 6H₂O. Twelve H atoms enter, six remain in the hydroxide and six form 3H₂.
2Al(s) + 6H₂O(l) → 2Al(OH)₃(s) + 3H₂(g)
| Element | Left | Right |
|---|---|---|
| Al | 2 | 2 |
| O | 6 | 2 × 3 = 6 |
| H | 6 × 2 = 12 | (2 × 3) + (3 × 2) = 12 |
This equation assumes the protective oxide layer has been overcome. Aluminium in ordinary cold water is protected, so do not predict immediate vigorous bubbling from the balanced equation alone.
6. Worked example 3 — phosphorus and chlorine
Task: Use P₄ and Cl₂ to make PCl₅. Four phosphorus atoms require 4PCl₅. Those four molecules/formula units contain 20 chlorine atoms, requiring 10Cl₂.
P₄(s) + 10Cl₂(g) → 4PCl₅(s)
Check P 4 = 4 and Cl 20 = 20. PCl₅ is the specified product with sufficient chlorine. If an aluminium equation uses Al₂Cl₆ instead, balance 2Al + 3Cl₂ → Al₂Cl₆; do not mix the dimer formula and monomer coefficient.
7. C1.5 acid–base equation bank
Choose the actual salt before balancing. Complete neutralisation of phosphoric acid gives phosphate; partial neutralisation gives hydrogen phosphates. The equations below specify complete neutralisation where relevant. Aqueous ionic equations omit spectator ions such as Na⁺ and Cl⁻.
| Type | Balanced equation | Key idea |
|---|---|---|
| Basic oxides | Na₂O(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) | Two H⁺ per oxide oxygen |
| Basic oxides | MgO(s) + 2HCl(aq) → MgCl₂(aq) + H₂O(l) | Salt and water, not hydrogen gas |
| Basic hydroxides | NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) | 1:1 neutralisation |
| Basic hydroxides | Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l) | Two OH groups need two H⁺ |
| Amphoteric oxide with acid | Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l) | Charge +6 on both sides |
| Amphoteric hydroxide with acid | Al(OH)₃(s) + 3H⁺(aq) → Al³⁺(aq) + 3H₂O(l) | Charge +3 on both sides |
| Amphoteric hydroxide with alkali | Al(OH)₃(s) + OH⁻(aq) → [Al(OH)₄]⁻(aq) | Excess strong aqueous alkali |
| Amphoteric oxide with alkali | Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq) | Strong aqueous alkali; heating may be needed |
| Acidic oxide | SiO₂(s) + 2NaOH(aq) → Na₂SiO₃(aq) + H₂O(l) | Hot concentrated aqueous alkali; simplified silicate representation |
| Acidic oxide | P₄O₁₀(s) + 12NaOH(aq) → 4Na₃PO₄(aq) + 6H₂O(l) | Enough base for phosphate |
| Acidic oxide | SO₂(g) + 2NaOH(aq) → Na₂SO₃(aq) + H₂O(l) | Enough base for sulfite |
| Acid | H₃PO₄(aq) + 3NaOH(aq) → Na₃PO₄(aq) + 3H₂O(l) | Three acidic protons |
| Acid | H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l) | Two acidic protons |
| Acids | HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) | Monoprotic |
| Acids | HClO(aq) + NaOH(aq) → NaClO(aq) + H₂O(l) | Retain ClO⁻ in the salt |
| Acids | HClO₄(aq) + NaOH(aq) → NaClO₄(aq) + H₂O(l) | Retain ClO₄⁻ in the salt |
8. Worked example 4 — phosphorus(V) oxide and alkali
Task: Balance P₄O₁₀ + NaOH → Na₃PO₄ + H₂O for full neutralisation. Four phosphorus atoms require 4Na₃PO₄. This requires 12 sodium atoms, so use 12NaOH. Twelve hydrogen atoms then form 6H₂O. Finally check oxygen: 10 + 12 = 22 on the left; 16 + 6 = 22 on the right.
P₄O₁₀(s) + 12NaOH(aq) → 4Na₃PO₄(aq) + 6H₂O(l)
The coefficients are 1:12:4:6. Do not simplify them by changing P₄O₁₀ into a different displayed formula when the question specifies P₄O₁₀.
9. Worked example 5 — an ionic amphoteric reaction
Task: Balance Al₂O₃ + OH⁻ + H₂O → [Al(OH)₄]⁻. Two aluminium atoms require two complex ions. Their total charge is −2, so use 2OH⁻. The products contain eight O and eight H atoms. The oxide and two hydroxides supply five O and two H, so add 3H₂O.
Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq)
| Check | Left | Right |
|---|---|---|
| Al | 2 | 2 |
| O | 3 + 2 + 3 = 8 | 2 × 4 = 8 |
| H | 2 + 6 = 8 | 2 × 4 = 8 |
| Charge | −2 | −2 |
Charge conservation is an additional check, not a substitute for checking the atoms. No electrons are needed in this acid–base equation.
10. C1.6 equations: dissolution and hydrolysis
| Situation | Balanced representation |
|---|---|
| NaCl dissolves | NaCl(s) → Na⁺(aq) + Cl⁻(aq) |
| MgCl₂ dissolves | MgCl₂(s) → Mg²⁺(aq) + 2Cl⁻(aq) |
| Weak magnesium aquo-ion hydrolysis | [Mg(H₂O)₆]²⁺(aq) + H₂O(l) ⇌ [Mg(H₂O)₅(OH)]⁺(aq) + H₃O⁺(aq) |
| Aluminium chloride: hydration accounting in excess water | Al₂Cl₆ + 12H₂O → 2[Al(H₂O)₆]³⁺ + 6Cl⁻ |
| Aluminium aquo-ion hydrolysis | [Al(H₂O)₆]³⁺(aq) + H₂O(l) ⇌ [Al(H₂O)₅(OH)]²⁺(aq) + H₃O⁺(aq) |
| SiCl₄, simplified overall hydrolysis | SiCl₄(l) + 2H₂O(l) → SiO₂(s) + 4HCl(aq) |
| PCl₅, limited water | PCl₅(s) + H₂O(l) → POCl₃(l) + 2HCl(g) |
| Further hydrolysis | POCl₃(l) + 3H₂O(l) → H₃PO₄(aq) + 3HCl(aq) |
| PCl₅, excess water overall | PCl₅(s) + 4H₂O(l) → H₃PO₄(aq) + 5HCl(aq) |
In the aluminium hydration accounting equation, starting phase depends on the stated conditions and the ions are aqueous. HCl may be written (g) when it escapes with limited water, or (aq) when it dissolves. Actual silicon hydrolysis commonly gives hydrated silica, represented by SiO₂ in the simplified equation. Keep these condition-dependent meanings attached to the equation.
11. Worked example 6 — combine the PCl₅ stages
Task: Derive the excess-water equation. Write the limited-water step and the further hydrolysis step, then cancel the intermediate POCl₃, which appears on both sides.
PCl₅ + H₂O → POCl₃ + 2HCl
POCl₃ + 3H₂O → H₃PO₄ + 3HCl
Adding gives 4H₂O as a reactant and 5HCl as a product:
PCl₅(s) + 4H₂O(l) → H₃PO₄(aq) + 5HCl(aq)
| Element | Left | Right |
|---|---|---|
| P | 1 | 1 |
| Cl | 5 | 5 |
| O | 4 | 4 |
| H | 8 | 3 + 5 = 8 |
The phosphorus oxidation number stays +5. Balanced hydrolysis does not require inserting electrons or calling it a redox reaction.
12. Independent practice
- Balance Al + O₂ → Al₂O₃, keeping all formulas unchanged.
- Balance SiCl₄ + H₂O → SiO₂ + HCl.
- Write complete neutralisation of H₂SO₄ with NaOH.
- Write the equation for Na₂O reacting with water.
Reveal practice answers
4Al + 3O₂ → 2Al₂O₃
SiCl₄ + 2H₂O → SiO₂ + 4HCl
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Na₂O + H₂O → 2NaOH
Add states based on the question: the usual room-temperature products are Al₂O₃(s), SiO₂(s), dissolved HCl, Na₂SO₄(aq), NaOH(aq) and H₂O(l).
Common misconceptions
- “Changing subscripts is balancing.” It changes the identity of the reactant or product.
- “Oxygen and chlorine are single atoms in the reactants.” Use O₂ and Cl₂ in these reactions.
- “A balanced equation must occur rapidly.” Passivation and conditions still control observed reactions.
- “All chloride–water equations form hydroxide and HCl.” NaCl mainly dissolves, hydrated metal ions establish equilibria, and covalent chlorides hydrolyse differently.
- “Only atoms matter in an ionic equation.” Total electrical charge must also balance.
Exam tips
- Write the correct product formulas before adjusting coefficients.
- Show an atom tally for complex equations and a charge tally for ionic ones.
- Read “excess water”, “limited water”, “excess alkali” and the specified oxide/chloride carefully.
- Use the smallest whole-number ratio and include state symbols when requested or chemically useful.
- Do not insert H₂ into an oxide–acid neutralisation; salt and water are the usual products.
Check your understanding
Select one option and Check Answer. Read the explanation, then use Try Again to reset.