When 3 moles of hydrogen gas (H_2) react completely with 3 moles of oxygen gas (O_2) according to the equation 2H_2 + O_2 → 2H_2O, how many grams of water (H_2O) are produced?

• A. 90 grams
• B. 36 grams
• C. 108 grams
• D. 54 grams

Answer: (C)

Verified step by step guidance

1. Write down the balanced chemical equation: \[2H_2 + O_2 \rightarrow 2H_2O\].
2. Identify the mole ratio between hydrogen gas and water from the balanced equation. For every 2 moles of \[H_2\], 2 moles of \[H_2O\] are produced, so the ratio is 1:1.
3. Determine the limiting reactant by comparing the given moles of \[H_2\] and \[O_2\] with the stoichiometric ratio. Since 3 moles of \[H_2\] require 1.5 moles of \[O_2\] (based on the ratio 2:1), and 3 moles of \[O_2\] are available, \[H_2\] is the limiting reactant.
4. Calculate the moles of water produced, which will be equal to the moles of the limiting reactant \[H_2\] (3 moles) because of the 1:1 mole ratio.
5. Calculate the mass of water produced by multiplying the moles of \[H_2O\] by its molar mass: \[\text{mass} = \text{moles} \times \text{molar mass} = 3 \times 18 \text{ g/mol}\].