According to the law of conservation of mass, if 32.0 g of oxygen gas reacts completely with methane (CH4) to produce carbon dioxide and water, what mass of methane is required for the reaction: CH4 + 2 O2 → CO2 + 2 H2O?

• A. 16.0 g
• B. 32.0 g
• C. 44.0 g
• D. 64.0 g

Answer: (A)

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. Identify the given mass of oxygen gas (\(\mathrm{O_2}\)), which is 32.0 g, and note that you need to find the mass of methane (\(\mathrm{CH_4}\)) required.
3. Calculate the molar mass of oxygen gas (\(\mathrm{O_2}\)): \$2 \times 16.00\ \mathrm{g/mol} = 32.00\ \mathrm{g/mol}$.
4. Determine the number of moles of oxygen gas reacting using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{32.0\ \mathrm{g}}{32.00\ \mathrm{g/mol}}\).
5. Use the mole ratio from the balanced equation to find moles of methane: for every 2 moles of \(\mathrm{O_2}\), 1 mole of \(\mathrm{CH_4}\) reacts. Calculate moles of \(\mathrm{CH_4}\) as \(\frac{1}{2}\) times the moles of \(\mathrm{O_2}\).
6. Calculate the molar mass of methane (\(\mathrm{CH_4}\)): \$12.01\ \mathrm{g/mol} + 4 \times 1.008\ \mathrm{g/mol}$.
7. Finally, find the mass of methane required by multiplying the moles of methane by its molar mass: \(\text{mass} = \text{moles} \times \text{molar mass}\).