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. 12
• B. 10
• C. 8
• D. 13

Answer: (D)

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 carbon atoms are calculated as: \$2 \times 4 = 8$ carbon atoms.
2. Check the product side for carbon atoms. The equation shows 8 CO_2 molecules, and each CO_2 contains 1 carbon atom, so the carbon atoms are balanced: 8 on both sides.
3. Next, count the hydrogen atoms on the reactant side. Each C_4H_{10} molecule has 10 hydrogen atoms, so total hydrogens are: \$2 \times 10 = 20$ hydrogen atoms.
4. On the product side, hydrogen atoms are in H_2O molecules. Since there are 10 H_2O molecules, and each has 2 hydrogen atoms, total hydrogens are: \$10 \times 2 = 20$, so hydrogens are balanced.
5. Now, balance the oxygen atoms. On the product side, oxygen atoms come from 8 CO_2 molecules and 10 H_2O molecules. Total oxygen atoms are: \((8 \times 2) + (10 \times 1) = 16 + 10 = 26\). Since each O_2 molecule has 2 oxygen atoms, the number of O_2 molecules needed is \(\frac{26}{2} = 13\). Therefore, the coefficient for O_2 is 13.