Grade 10 Science · Problem type

Balancing a synthesis: aluminum oxide

Use this for “Balance Al + O₂ → Al₂O₃.”

Diagram

AlAlAlAl4 Al+OOOOOO3 O₂OAlOAlOOAlOAlO2 Al₂O₃BeforeAfterO6=6Al4=4
4 Al + 3 O₂ → 2 Al₂O₃
Oxygen: 3 × 2 = 6 before, 2 × 3 = 6 after
Aluminum: 4 × 1 = 4 before, 2 × 2 = 4 after
Balanced: the same atoms on both sides, only rearranged.
This is an example. Choose your own values to start a new problem.
a
Al
+
b
O₂
→
c
Al₂O₃

Formulas

c
=
a
2
2
=
4
2
b
=
3
×
c
2
3
=
3
×
2
2

Assumptions

•
Aluminum burns in oxygen to make aluminum oxide, one product from two reactants.
•
Oxygen comes in pairs and Al₂O₃ holds 3, so the oxygen atoms must be a multiple of 6: 4 aluminum atoms at a time.
•
The balanced equation uses the smallest whole numbers, 4, 3 and 2; larger inputs are the same reaction run more times.
•
Aluminum oxide is ionic: each formula unit is two Al³⁺ and three O²⁻, drawn as ions with no bonds.

Step-by-step

WE KNOW
a = 4
FIND
aluminum oxide formula units (c), oxygen molecules (b)
Step 1 · Find aluminum oxide formula units (c)
Use c = a/2
Aluminum: each Al₂O₃ holds 2 aluminum atoms, so half as many formula units as atoms.
c
=
4
2
c
=
2
Step 2 · Find oxygen molecules (b)
Use b = 3 × c/2
Oxygen: each Al₂O₃ holds 3 oxygen atoms and each O₂ brings 2.
b
=
3
×
2
2
b
=
3
Check
Put the numbers back in. Both sides should match.
2
=
4
2
✓
3
=
3
×
2
2
✓
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