In the reaction 2 H_2 + O_2 → 2 H_2O, how many grams of oxygen gas (O_2) are required to produce 36.0 grams of water (H_2O)?

• A. 36.0 grams
• B. 18.0 grams
• C. 32.0 grams
• D. 64.0 grams

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \$2\ H_2 + O_2 \rightarrow 2\ H_2O$.
2. Calculate the molar mass of water (\(H_2O\)). Use atomic masses: H = 1.0 g/mol, O = 16.0 g/mol, so \(M_{H_2O} = 2 \times 1.0 + 16.0 = 18.0\ \text{g/mol}\).
3. Convert the given mass of water (36.0 g) to moles using the molar mass: \(\text{moles of } H_2O = \frac{36.0\ \text{g}}{18.0\ \text{g/mol}}\).
4. Use the stoichiometric ratio from the balanced equation to find moles of oxygen gas (\(O_2\)) needed. According to the equation, 2 moles of \(H_2O\) are produced per 1 mole of \(O_2\), so \(\text{moles of } O_2 = \frac{1}{2} \times \text{moles of } H_2O\).
5. Calculate the mass of oxygen gas required by multiplying the moles of \(O_2\) by its molar mass (\(M_{O_2} = 2 \times 16.0 = 32.0\ \text{g/mol}\)): \(\text{mass of } O_2 = \text{moles of } O_2 \times 32.0\ \text{g/mol}\).