SME
Grade 10 Science · Problem type
Balancing the combustion of an alkane
Use this to balance any alkane burning, such as “C₃H₈ + O₂ → CO₂ + H₂O”: carbon first, hydrogen next, oxygen last.
Diagram
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Carbon: 1 × 3 = 3 before, 3 × 1 = 3 after
Hydrogen: 1 × 8 = 8 before, 4 × 2 = 8 after
Oxygen: 5 × 2 = 10 before, 3 × 2 + 4 × 1 = 10 after
Balanced: the same atoms on both sides, only rearranged.
This is an example. Type your own values to start a new problem.
a
C
H
+
b
O₂
→
c
CO₂
+
d
H₂O
A coefficient of 1 is not written: 1 O₂ is written O₂.
Formulas
y
=
2
×
x
+
2
8 = 2 × 3 + 2
a
=
1
if
y
is
a
multiple
of
4,
else
2
1 = 1 if 8 is a multiple of 4, else 2
c
=
a
×
x
3 = 1 × 3
d
=
a
×
y
2
4
=
1
×
8
2
b
=
c
+
d
2
5
=
3
+
4
2
Assumptions
•
A hydrocarbon burns in oxygen to make carbon dioxide and water; coefficients count molecules, subscripts never change.
•
An alkane has 2x + 2 hydrogen atoms for x carbon atoms.
•
Balance carbon first, then hydrogen, and oxygen last, since O₂ appears alone.
•
When the hydrogens are not a multiple of 4, the O₂ would come out as a half: double the fuel.
Step-by-step
WE KNOW
x = 3
FIND
hydrogen atoms in the fuel (y), fuel molecules (a), carbon dioxide molecules (c), water molecules (d), oxygen molecules (b)
Step 1 · Find hydrogen atoms in the fuel (y)
Use y = 2 × x + 2
Each carbon holds 2 hydrogens, and the two ends 1 more each.
y
=
2
×
3
+
2
y
=
6
+
2
y
=
8
Step 2 · Find fuel molecules (a)
Use a = 1 if y is a multiple of 4, else 2
Each O₂ brings 2 oxygen atoms. With one fuel molecule the water’s oxygen would leave half an O₂ unless the hydrogens are a multiple of 4.
a
=
1
if
8
is
a
multiple
of
4,
else
2
a
=
1
Step 3 · Find carbon dioxide molecules (c)
Use c = a × x
Carbon first: one CO₂ for each carbon atom in the fuel.
c
=
1
×
3
c
=
3
Step 4 · Find water molecules (d)
Use d = a × y/2
Hydrogen next: each H₂O takes 2 of the fuel’s hydrogen atoms.
d
=
1
×
8
2
d
=
4
Step 5 · Find oxygen molecules (b)
Use b = c + d/2
Oxygen last: 2 atoms in each CO₂ and 1 in each H₂O, halved because each O₂ brings 2.
b
=
3
+
4
2
6 + 4 = 10 oxygen atoms on the right, so 5 O₂ on the left
b
=
5
Check
Put the numbers back in. Both sides should match.
8 = 2 × 3 + 2 ✓
1 = 1 if 8 is a multiple of 4, else 2 ✓
3 = 1 × 3 ✓
4
=
1
×
8
2
✓
5
=
3
+
4
2
✓
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