In terms of stoichiometry, what is the coefficient of O_2 in the balanced chemical equation for the combustion of C_2H_6?

• A. 2
• B. 3
• C. 7
• D. 3.5

Answer: (B)

Verified step by step guidance

1. Write the unbalanced chemical equation for the combustion of ethane (C_2H_6): \(\mathrm{C_2H_6} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}\).
2. Balance the carbon atoms first. Since there are 2 carbon atoms in C_2H_6, place a coefficient of 2 in front of CO_2: \(\mathrm{C_2H_6} + \mathrm{O_2} \rightarrow 2\mathrm{CO_2} + \mathrm{H_2O}\).
3. Balance the hydrogen atoms next. There are 6 hydrogen atoms in C_2H_6, so place a coefficient of 3 in front of H_2O (because each water molecule has 2 hydrogens): \(\mathrm{C_2H_6} + \mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O}\).
4. Balance the oxygen atoms last. Count the total oxygen atoms on the right side: \$2 \times 2 = 4\( from CO_2 and \)3 \times 1 = 3\( from H_2O, totaling 7 oxygen atoms. Since O_2 is diatomic, divide 7 by 2 to find the coefficient for O_2: \)\frac{7}{2} = 3.5$.
5. To avoid fractional coefficients, multiply all coefficients by 2 to get whole numbers: \$2 \times \mathrm{C_2H_6} + 7 \times \mathrm{O_2} \rightarrow 4 \times \mathrm{CO_2} + 6 \times \mathrm{H_2O}$. Thus, the coefficient of O_2 is 7.