In the reaction N_2 + 3H_2 → 2NH_3, how many moles of nitrogen gas (N_2) are required to completely react with 6.34 mol of hydrogen gas (H_2)?

• A. 6.34 mol
• B. 3.17 mol
• C. 2.11 mol
• D. 1.06 mol

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation: \(\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}\).
2. Note the mole ratio between nitrogen gas (\(\mathrm{N_2}\)) and hydrogen gas (\(\mathrm{H_2}\)) from the balanced equation, which is 1 mole of \(\mathrm{N_2}\) reacts with 3 moles of \(\mathrm{H_2}\).
3. Set up a proportion to find the moles of \(\mathrm{N_2}\) needed to react with 6.34 moles of \(\mathrm{H_2}\) using the mole ratio: \(\mathrm{\frac{1\ mol\ N_2}{3\ mol\ H_2} = \frac{x\ mol\ N_2}{6.34\ mol\ H_2}}\).
4. Solve for \(x\) (the moles of \(\mathrm{N_2}\)) by cross-multiplying and dividing: \(x = \frac{1}{3} \times 6.34\) moles.
5. Interpret the result as the number of moles of nitrogen gas required to completely react with 6.34 moles of hydrogen gas.