Which of the following is the correctly balanced chemical equation for the combustion of propane (C_3H_8)?

• A. C_3H_8 + 5 O_2 → 2 CO_2 + 4 H_2O
• B. C_3H_8 + 3 O_2 → 3 CO_2 + 4 H_2O
• C. C_3H_8 + 4 O_2 → 3 CO_2 + 4 H_2O
• D. C_3H_8 + 5 O_2 → 3 CO_2 + 4 H_2O

Answer: (D)

Verified step by step guidance

1. Write the unbalanced combustion reaction of propane: \(\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}\).
2. Balance the carbon atoms first. Since propane has 3 carbon atoms, place a coefficient of 3 in front of \(\mathrm{CO_2}\): \(\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow 3 \mathrm{CO_2} + \mathrm{H_2O}\).
3. Balance the hydrogen atoms next. Propane has 8 hydrogen atoms, so place a coefficient of 4 in front of \(\mathrm{H_2O}\) because each water molecule has 2 hydrogens: \(\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow 3 \mathrm{CO_2} + 4 \mathrm{H_2O}\).
4. Count the total oxygen atoms on the product side: \$3 \times 2 = 6\( from \)\mathrm{CO_2}\( and \)4 \times 1 = 4\( from \)\mathrm{H_2O}$, totaling 10 oxygen atoms.
5. Balance the oxygen atoms on the reactant side by placing a coefficient of 5 in front of \(\mathrm{O_2}\) (since each \(\mathrm{O_2}\) molecule has 2 oxygen atoms): \(\mathrm{C_3H_8} + 5 \mathrm{O_2} \rightarrow 3 \mathrm{CO_2} + 4 \mathrm{H_2O}\).