Given the reaction N2 + 3 H2 → 2 NH3, how many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?

• A. 21.0 mol
• B. 14.0 mol
• C. 7.0 mol
• D. 42.0 mol

Answer: (B)

Verified step by step guidance

1. Identify the balanced chemical equation: \(\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}\).
2. Determine the mole ratio between \(\mathrm{H_2}\) and \(\mathrm{NH_3}\) from the balanced equation. For every 3 moles of \(\mathrm{H_2}\), 2 moles of \(\mathrm{NH_3}\) are produced.
3. Since \(\mathrm{N_2}\) is in excess, the amount of \(\mathrm{H_2}\) limits the reaction. Use the mole ratio to calculate moles of \(\mathrm{NH_3}\) produced from 21.0 moles of \(\mathrm{H_2}\).
4. Set up the conversion: \(\mathrm{21.0\ mol\ H_2 \times \frac{2\ mol\ NH_3}{3\ mol\ H_2}}\) to find moles of \(\mathrm{NH_3}\).
5. Perform the calculation to find the number of moles of \(\mathrm{NH_3}\) produced.