At STP, how many liters of O_2 are required to produce 2 liters of H_2O(g) according to the reaction: 2H_2(g) + O_2(g) → 2H_2O(g)?

• A. 2 liters
• B. 0.5 liter
• C. 4 liters
• D. 1 liter

Answer: (D)

Verified step by step guidance

1. Write down the balanced chemical equation: \$2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$.
2. Identify the mole ratio between \(O_2\) and \(H_2O\) from the balanced equation. For every 2 moles of \(H_2O\) produced, 1 mole of \(O_2\) is consumed.
3. Recall that at STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters, and gas volumes in reactions relate directly to mole ratios when measured at the same conditions.
4. Use the volume ratio from the balanced equation: since 2 liters of \(H_2O\) are produced, the volume of \(O_2\) required is half that amount, based on the 1:2 ratio.
5. Calculate the volume of \(O_2\) needed by applying the ratio: \(V_{O_2} = \frac{1}{2} \times V_{H_2O}\), where \(V_{H_2O} = 2\) liters.