Given the reaction NBr3 + 3 NaOH → 3 NaBr + N(OH)3, if there are 40 mol of NBr3 and 48 mol of NaOH, what is the limiting reactant?

• A. Both are limiting
• B. NaOH
• C. Neither is limiting
• D. NBr3

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: \(\mathrm{NBr_3 + 3\ NaOH \rightarrow 3\ NaBr + N(OH)_3}\).
2. Identify the mole ratio between the reactants from the balanced equation: 1 mole of \(\mathrm{NBr_3}\) reacts with 3 moles of \(\mathrm{NaOH}\).
3. Calculate the amount of \(\mathrm{NaOH}\) required to completely react with 40 moles of \(\mathrm{NBr_3}\) using the mole ratio: \$40\ \mathrm{mol\ NBr_3} \times \frac{3\ \mathrm{mol\ NaOH}}{1\ \mathrm{mol\ NBr_3}}$.
4. Compare the calculated required moles of \(\mathrm{NaOH}\) with the available 48 moles to determine if \(\mathrm{NaOH}\) is sufficient or limiting.
5. If the available \(\mathrm{NaOH}\) is less than the required amount, then \(\mathrm{NaOH}\) is the limiting reactant; otherwise, \(\mathrm{NBr_3}\) is limiting.