When 8 grams of H_2 gas reacts completely with excess O_2 at standard temperature and pressure (STP), what volume of water vapor (H_2O) is produced?

• A. 11.2 L
• B. 22.4 L
• C. 44.8 L
• D. 72.0 L

Answer: (D)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between hydrogen gas and oxygen gas to form water vapor: \$2H_2(g) + O_2(g) \rightarrow 2H_2O(g)$.
2. Calculate the number of moles of \(H_2\) gas reacting using its molar mass. Use the formula: \(\text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2}\), where the molar mass of \(H_2\) is approximately 2 g/mol.
3. Use the stoichiometric ratio from the balanced equation to find the moles of \(H_2O\) produced. According to the equation, 2 moles of \(H_2\) produce 2 moles of \(H_2O\), so the moles of \(H_2O\) will be equal to the moles of \(H_2\).
4. At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters. Use this to calculate the volume of water vapor produced: \(\text{Volume of } H_2O = \text{moles of } H_2O \times 22.4 \text{ L/mol}\).
5. Substitute the values calculated in previous steps into the volume formula to find the volume of water vapor produced.