Given the reaction 2 H_2 + O_2 → 2 H_2O, what mass of O_2 (in grams) is required to produce 55.0 grams of H_2O?

• A. 32.0 g
• B. 16.0 g
• C. 48.7 g
• D. 22.0 g

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation: \$2\ H_2 + O_2 \rightarrow 2\ H_2O$. This tells us the mole ratio between reactants and products.
2. Calculate the number of moles of \(H_2O\) produced using its molar mass. Use the formula: \(\text{moles of } H_2O = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O}\), where the molar mass of \(H_2O\) is approximately 18.0 g/mol.
3. Use the mole ratio from the balanced equation to find the moles of \(O_2\) required. According to the equation, 2 moles of \(H_2O\) are produced per 1 mole of \(O_2\), so \(\text{moles of } O_2 = \frac{1}{2} \times \text{moles of } H_2O\).
4. Calculate the mass of \(O_2\) needed by multiplying the moles of \(O_2\) by its molar mass. Use the formula: \(\text{mass of } O_2 = \text{moles of } O_2 \times \text{molar mass of } O_2\), where the molar mass of \(O_2\) is approximately 32.0 g/mol.
5. Summarize the process: convert given mass of \(H_2O\) to moles, use mole ratio to find moles of \(O_2\), then convert moles of \(O_2\) to mass. This will give the mass of \(O_2\) required to produce 55.0 grams of \(H_2O\).