In the balanced chemical equation for the complete combustion of butane (C_4H_{10}), what is the mole ratio of butane (C_4H_{10}) to carbon dioxide (CO_2)?

• A. 1:2
• B. 4:8
• C. 1:4
• D. 2:8

Answer: (D)

Verified step by step guidance

1. Write the unbalanced chemical equation for the complete combustion of butane: \(\mathrm{C_4H_{10} + O_2 \rightarrow CO_2 + H_2O}\).
2. Balance the carbon atoms first. Since butane has 4 carbon atoms, place a coefficient of 4 in front of \(\mathrm{CO_2}\): \(\mathrm{C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O}\).
3. Balance the hydrogen atoms next. Butane has 10 hydrogen atoms, so place a coefficient of 5 in front of \(\mathrm{H_2O}\) (because each water molecule has 2 hydrogens): \(\mathrm{C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O}\).
4. Balance the oxygen atoms last. On the right side, there are \$4 \times 2 = 8\( oxygen atoms from \)\mathrm{CO_2}\( and \)5 \times 1 = 5\( oxygen atoms from \)\mathrm{H_2O}\(, totaling 13 oxygen atoms. Since oxygen gas is \)\mathrm{O_2}\(, divide 13 by 2 to get the coefficient for \)\mathrm{O_2}\(: \)\frac{13}{2}$.
5. To avoid fractional coefficients, multiply the entire equation by 2, resulting in the balanced equation: \(\mathrm{2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O}\). From this, the mole ratio of butane to carbon dioxide is \$2:8$.