If 4.00 g of hydrogen gas (H_2) reacts completely with excess oxygen gas (O_2) to form water (H_2O), how many grams of water are produced? (Molar masses: H_2 = 2.02 g/mol, H_2O = 18.02 g/mol)

• A. 36.0 g
• B. 18.0 g
• C. 4.0 g
• D. 9.0 g

Answer: (B)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction: \(\mathrm{2H_2 + O_2 \rightarrow 2H_2O}\).
2. Calculate the number of moles of hydrogen gas (\(\mathrm{H_2}\)) using its molar mass: \(\text{moles of } H_2 = \frac{\text{mass of } H_2}{\text{molar mass of } H_2} = \frac{4.00\ \mathrm{g}}{2.02\ \mathrm{g/mol}}\).
3. Use the mole ratio from the balanced equation to find the moles of water (\(\mathrm{H_2O}\)) produced. According to the equation, 2 moles of \(\mathrm{H_2}\) produce 2 moles of \(\mathrm{H_2O}\), so moles of \(\mathrm{H_2O}\) = moles of \(\mathrm{H_2}\).
4. Calculate the mass of water produced by multiplying the moles of water by the molar mass of water: \(\text{mass of } H_2O = \text{moles of } H_2O \times 18.02\ \mathrm{g/mol}\).
5. Interpret the result as the mass of water formed when 4.00 g of hydrogen gas reacts completely with excess oxygen.