Balance the combustion of propane: C3H8 + O2 → CO2 + H2O; provide the balanced coefficients.

• A. C3H8 + 5 O2 -> 3 CO2 + 4 H2O
• B. C3H8 + O2 -> CO2 + H2O
• C. 2 C3H8 + 7 O2 -> 6 CO2 + 8 H2O
• D. C3H8 + 3 O2 -> 2 CO2 + 3 H2O

Answer: (A)

Balancing combustion hinges on conserving each type of atom. Propane has 3 carbon atoms and 8 hydrogen atoms, so the products must include 3 carbon dioxide and 4 water molecules to match those counts. That gives C3H8 + ? O2 → 3 CO2 + 4 H2O. Now count oxygen atoms: the right side has 3 CO2 (9 O) and 4 H2O (4 O), totaling 13 O, but remember each CO2 has 2 O and each H2O has 1 O, so the right side actually has 3×2 + 4×1 = 6 + 4 = 10 O atoms. To supply 10 O atoms on the left, you need 5 O2 molecules (since each O2 has 2 O atoms, 5×2 = 10). Thus the balanced equation is C3H8 + 5 O2 → 3 CO2 + 4 H2O. The other proposed sets fail to balance carbon, hydrogen, or oxygen counts, so they do not conserve atoms.