Balancing Chemical Equations

This lesson teaches you the systematic method for balancing chemical equations as required by the Edexcel GCSE Combined Science specification (1SC0). You will practise writing balanced symbol equations with state symbols and develop confidence with a wide range of equation types.

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Why Must Equations Be Balanced?

Because of the law of conservation of mass:

• The same number of each type of atom must appear on both sides of the equation.
• We balance equations by placing numbers (coefficients) in front of formulae — we NEVER change the formulae themselves.

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State Symbols

Every balanced equation should include state symbols to show the physical state of each substance:

Symbol	Meaning
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous (dissolved in water)

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The Systematic Method for Balancing

Follow these steps every time:

Step 1: Write the correct formulae for all reactants and products (do NOT change these).

Step 2: Count the number of each type of atom on both sides.

Step 3: Balance one element at a time by adjusting coefficients.

Step 4: Start with elements that appear in only one compound on each side.

Step 5: Leave elements that appear in many places (often H or O) until last.

Step 6: Check every atom balances.

Step 7: Add state symbols.

flowchart TD
    A[Write correct formulae for reactants and products] --> B[Count atoms of each element on both sides]
    B --> C{Atoms balanced?}
    C -- No --> D[Adjust coefficient of one species]
    D --> E[Start with metals or unique elements]
    E --> F[Leave H and O until last]
    F --> B
    C -- Yes --> G[Add state symbols s, l, g, aq]
    G --> H[Final balanced equation]

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Worked Examples

Example 1 — Hydrogen + Oxygen → Water

Unbalanced: $\text{H}_2 + \text{O}_2 \rightarrow \text{H}_2\text{O}$H2​+O2​→H2​O

Atom	Left	Right
H	2	2
O	2	1 ← unbalanced

Put a 2 in front of $\text{H}_2\text{O}$H2​O:

$\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$H2​+O2​→2H2​O

Atom	Left	Right
H	2	4 ← unbalanced
O	2	2 ✓

Put a 2 in front of $\text{H}_2$H2​:

$2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$2H2​(g)+O2​(g)→2H2​O(l)

Atom	Left	Right
H	4	4 ✓
O	2	2 ✓

Example 2 — Sodium + Water → Sodium Hydroxide + Hydrogen

Unbalanced: $\text{Na} + \text{H}_2\text{O} \rightarrow \text{NaOH} + \text{H}_2$Na+H2​O→NaOH+H2​

Atom	Left	Right
Na	1	1 ✓
H	2	3 ← unbalanced
O	1	1 ✓

Put a 2 in front of $\text{Na}$Na, $\text{H}_2\text{O}$H2​O, and $\text{NaOH}$NaOH:

$2\text{Na}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq) + \text{H}_2(g)$2Na(s)+2H2​O(l)→2NaOH(aq)+H2​(g)

Atom	Left	Right
Na	2	2 ✓
H	4	2 + 2 = 4 ✓
O	2	2 ✓

Example 3 — Iron + Oxygen → Iron(III) Oxide

Unbalanced: $\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$Fe+O2​→Fe2​O3​

Atom	Left	Right
Fe	1	2
O	2	3

Balance Fe first — put 2 in front of Fe:

$2\text{Fe} + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3$2Fe+O2​→Fe2​O3​

Balance O — we need 3 on the left. Since O comes as $\text{O}_2$O2​, put 3 in front of $\text{O}_2$O2​ (giving 6 O) and 2 in front of $\text{Fe}_2\text{O}_3$Fe2​O3​ (giving 6 O):

$2\text{Fe} + 3\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3$2Fe+3O2​→2Fe2​O3​

Now Fe is unbalanced: 2 on left, 4 on right. Put 4 in front of Fe:

$4\text{Fe}(s) + 3\text{O}_2(g) \rightarrow 2\text{Fe}_2\text{O}_3(s)$4Fe(s)+3O2​(g)→2Fe2​O3​(s)

Atom	Left	Right
Fe	4	4 ✓
O	6	6 ✓

Example 4 — Methane + Oxygen → Carbon Dioxide + Water

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$CH4​(g)+2O2​(g)→CO2​(g)+2H2​O(l)

Atom	Left	Right
C	1	1 ✓
H	4	4 ✓
O	4	2 + 2 = 4 ✓

Exam Tip: NEVER change the small (subscript) numbers in a formula when balancing. Only change the large (coefficient) numbers in front.

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Practice Equations

Try balancing these, then check the answers below:

1. $\text{Mg} + \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$Mg+HCl→MgCl2​+H2​
2. $\text{N}_2 + \text{H}_2 \rightarrow \text{NH}_3$N2​+H2​→NH3​
3. $\text{Al} + \text{O}_2 \rightarrow \text{Al}_2\text{O}_3$Al+O2​→Al2​O3​
4. $\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow \text{Fe} + \text{CO}_2$Fe2​O3​+CO→Fe+CO2​
5. $\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$C2​H6​+O2​→CO2​+H2​O

Answers

1. $\text{Mg}(s) + 2\text{HCl}(aq) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g)$Mg(s)+2HCl(aq)→MgCl2​(aq)+H2​(g)
2. $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$N2​(g)+3H2​(g)→2NH3​(g)
3. $4\text{Al}(s) + 3\text{O}_2(g) \rightarrow 2\text{Al}_2\text{O}_3(s)$4Al(s)+3O2​(g)→2Al2​O3​(s)
4. $\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)$Fe2​O3​(s)+3CO(g)→2Fe(s)+3CO2​(g)
5. $2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(l)$2C2​H6​(g)+7O2​(g)→4CO2​(g)+6H2​O(l)

Exam Tip: In the exam, you may be given a partially balanced equation and asked to find the missing coefficient. Use the atom count method to find it systematically.

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Common Mistakes to Avoid

Mistake	Correction
Changing subscripts in formulae	Only change coefficients (big numbers in front)
Forgetting state symbols	Always include (s), (l), (g), or (aq)
Not checking every atom at the end	Recount all atoms after balancing
Balancing O or H first	Start with metals/less common elements; leave O and H until last

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Summary

• Balanced equations have the same number of each type of atom on both sides.
• Only adjust coefficients — never change chemical formulae.
• Use the systematic method: count → balance one element at a time → check → add state symbols.
• Start with atoms that appear in only one place on each side.
• Leave hydrogen and oxygen until last.
• Always include state symbols: (s), (l), (g), (aq).

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Extended Worked Examples