Chemistry · General chemistry I · Worked example

Balance the combustion of propane

Balance the combustion of propane: C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(g).

C_{3} H_{8} + O_{2} \to {CO}_{2} + H_{2} O

Balance carbon

Propane has three carbon atoms, and CO₂ is the only product with carbon, so CO₂ gets a coefficient of 3.

C_{3} H_{8} + O_{2} \to 3 {CO}_{2} + H_{2} O

Balance hydrogen

Propane has eight hydrogen atoms, and each water molecule has two, so water gets a coefficient of 4.

C_{3} H_{8} + O_{2} \to 3 {CO}_{2} + 4 H_{2} O

Balance oxygen last

The products now hold 3 × 2 + 4 × 1 = 10 oxygen atoms. O₂ supplies two per molecule, so it needs a coefficient of 5. O₂ is left for last because it is a free element: changing it affects no other count.

C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {CO}_{2} (g) + 4 H_{2} O (g)

\begin{array}{c} C_{3} H_{8} (g) + 5 O_{2} (g) \\ \to 3 {CO}_{2} (g) + 4 H_{2} O (g) \end{array}

Count every element

Each element has the same count on both sides, so the equation is balanced. The coefficients 1 : 5 : 3 : 4 are the mole ratios a stoichiometry problem would use.

Atom count in the balanced equation
Element	Reactant side	Product side
C	3	3
H	8	8
O	10	10

Result

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g).

Your turn

Balance the combustion of ethane: C₂H₆ + O₂ → CO₂ + H₂O.

Show the answer and explanation

2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O.

Carbon: 2CO₂. Hydrogen: 3H₂O (6 atoms). Oxygen: 2 × 2 + 3 = 7 atoms, so 7/2 O₂. Doubling every coefficient clears the fraction.

2 C_{2} H_{6} + 7 O_{2} \to 4 {CO}_{2} + 6 H_{2} O

Keep exploring

Open the reaction in a Chemistry box and change the 5 to 4: the checker reports which element no longer balances.

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OpenStax Chemistry 2e — Writing and balancing chemical equations

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