How many moles of water are produced when 58 g of oxygen gas (O_2) react completely with excess hydrogen according to the reaction: 2 H_2 + O_2 → 2 H_2O?

• A. 3.6 moles
• B. 4.0 moles
• C. 1.8 moles
• D. 2.0 moles

Answer: (A)

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 oxygen gas (\(O_2\)). Since each oxygen atom has an atomic mass of approximately 16 g/mol, the molar mass of \(O_2\) is \$2 \times 16 = 32$ g/mol.
3. Determine the number of moles of \(O_2\) by dividing the given mass of oxygen gas by its molar mass: \(\text{moles of } O_2 = \frac{58\ \text{g}}{32\ \text{g/mol}}\).
4. Use the stoichiometric ratio from the balanced equation to find the moles of water produced. According to the equation, 1 mole of \(O_2\) produces 2 moles of \(H_2O\), so multiply the moles of \(O_2\) by 2: \(\text{moles of } H_2O = 2 \times \text{moles of } O_2\).
5. The result from the previous step gives the number of moles of water produced when 58 g of oxygen gas reacts completely with excess hydrogen.