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
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1
Write the unbalanced combustion reaction of propane: \(\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}\).
Balance the carbon atoms first. Since propane has 3 carbon atoms, place a coefficient of 3 in front of \(\mathrm{CO_2}\): \(\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow 3 \mathrm{CO_2} + \mathrm{H_2O}\).
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: \(\mathrm{C_3H_8} + \mathrm{O_2} \rightarrow 3 \mathrm{CO_2} + 4 \mathrm{H_2O}\).
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.
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): \(\mathrm{C_3H_8} + 5 \mathrm{O_2} \rightarrow 3 \mathrm{CO_2} + 4 \mathrm{H_2O}\).