Balance the combustion equation of the hydrocarbon C2H6 with O2 to form CO2 and H2O.

• A. 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
• B. C2H6 + O2 → CO2 + H2O
• C. C2H6 + 3 O2 → 2 CO2 + 3 H2O
• D. 2 C2H6 + 5 O2 → 2 CO2 + 6 H2O

Answer: (A)

Balancing hinges on conserving every atom: the number of carbon, hydrogen, and oxygen atoms on the left must equal those on the right. For the combustion of a hydrocarbon, balance carbon first by making the number of CO2 molecules match the carbon count in the fuel. Two molecules of ethane contain four carbons, so you need four CO2 molecules on the products side. Next, balance hydrogen by matching the hydrogens: eight? no, twelve hydrogens come from 2 C2H6, so you need six H2O molecules to supply 12 hydrogens. Now count oxygen on the product side: four CO2 give 8 oxygens, plus six H2O give 6 more, totaling 14 oxygens. Each O2 molecule provides 2 oxygens, so you need 7 O2 to supply 14 oxygens. Putting it together: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O. This satisfies balance for C, H, and O. The other forms fail because they misbalance at least one element (carbon, hydrogen, or oxygen) when you test the counts, so they don’t represent a correct balanced equation.