What is the correct coefficient for O_2(g) when balancing the following chemical equation? 2C_4H_{10}(g) + O_2(g) → 8CO_2(g) + 10H_2O(g)

• A. 13
• B. 8
• C. 10
• D. 12

Answer: (A)

Verified step by step guidance

1. Identify the number of carbon atoms on the reactant side. Since there are 2 molecules of C_4H_{10}, and each molecule contains 4 carbon atoms, the total number of carbon atoms is calculated as \$2 \times 4 = 8$.
2. Balance the carbon atoms by ensuring the same number of carbon atoms appear in the products. The product side has \$8CO_2$, which contains 8 carbon atoms, so carbon is balanced.
3. Balance the hydrogen atoms next. There are 2 molecules of C_4H_{10}, each with 10 hydrogen atoms, so total hydrogen atoms are \$2 \times 10 = 20\(. On the product side, water (\)H_2O\() contains 2 hydrogen atoms per molecule, so to balance hydrogen, you need \)10H_2O$ molecules.
4. Balance the oxygen atoms last. Count the oxygen atoms on the product side: \$8CO_2\( contributes \)8 \times 2 = 16\( oxygen atoms, and \)10H_2O\( contributes \)10 \times 1 = 10\( oxygen atoms, for a total of \)16 + 10 = 26$ oxygen atoms.
5. Since oxygen gas (\(O_2\)) is diatomic, each molecule contains 2 oxygen atoms. To balance 26 oxygen atoms, divide by 2 to find the coefficient for \(O_2\): \(\frac{26}{2} = 13\). Therefore, the coefficient for \(O_2\) is 13.