Given the reaction N2 + 3 H2 → 2 NH3, how many moles of NH3 can be produced from 16.5 mol of H2 and excess N2?

• A. 16.5 mol
• B. 11.0 mol
• C. 5.50 mol
• D. 8.25 mol

Answer: (B)

Verified step by step guidance

1. Identify the balanced chemical equation: \(\mathrm{N_2 + 3 H_2 \rightarrow 2 NH_3}\).
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. Use the given amount of \(\mathrm{H_2}\) (16.5 mol) and apply the mole ratio to find the moles of \(\mathrm{NH_3}\) produced: \(\mathrm{16.5\ mol\ H_2 \times \frac{2\ mol\ NH_3}{3\ mol\ H_2}}\).
4. Calculate the value from the above expression to find the moles of \(\mathrm{NH_3}\) produced (do not compute the final number here, just set up the calculation).
5. Since \(\mathrm{N_2}\) is in excess, it does not limit the reaction, so the calculated moles of \(\mathrm{NH_3}\) represent the maximum amount produced.