In the reaction CH_4 + 2 O_2 → CO_2 + 2 H_2O, how many grams of CO_2 are produced when 125 g of O_2 react with excess CH_4?

• A. 44.0 g
• B. 62.5 g
• C. 86.0 g
• D. 125 g

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{CH_4 + 2\ O_2 \rightarrow CO_2 + 2\ H_2O}\).
2. Calculate the molar mass of \(\mathrm{O_2}\). Since oxygen has an atomic mass of approximately 16.0 g/mol, \(\mathrm{O_2}\) has a molar mass of \$2 \times 16.0 = 32.0\ \mathrm{g/mol}$.
3. Convert the given mass of \(\mathrm{O_2}\) (125 g) to moles using the formula: \(\text{moles of } O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{125}{32.0}\).
4. Use the stoichiometric ratio from the balanced equation to find moles of \(\mathrm{CO_2}\) produced. According to the equation, 2 moles of \(\mathrm{O_2}\) produce 1 mole of \(\mathrm{CO_2}\), so \(\text{moles of } CO_2 = \frac{1}{2} \times \text{moles of } O_2\).
5. Calculate the mass of \(\mathrm{CO_2}\) produced by multiplying the moles of \(\mathrm{CO_2}\) by its molar mass (44.0 g/mol): \(\text{mass of } CO_2 = \text{moles of } CO_2 \times 44.0\).