To determine the excess reactant in the given chemical reaction:
[tex]\[ 2 NBr_3 + 3 NaOH \rightarrow N_2 + 3 NaBr + 3 HOBr \][/tex]
we need to follow these steps:
1. Identify the stoichiometric coefficients:
- For [tex]\(NBr_3\)[/tex]: 2
- For [tex]\(NaOH\)[/tex]: 3
2. Determine the moles of each reactant:
- [tex]\(NBr_3\)[/tex]: 40 moles
- [tex]\(NaOH\)[/tex]: 48 moles
3. Calculate the stoichiometric ratios:
- For [tex]\(NBr_3\)[/tex]: [tex]\(\frac{40}{2} = 20\)[/tex]
- For [tex]\(NaOH\)[/tex]: [tex]\(\frac{48}{3} = 16\)[/tex]
4. Compare the stoichiometric ratios to identify the limiting reactant:
- [tex]\(NBr_3\)[/tex] has a ratio of 20
- [tex]\(NaOH\)[/tex] has a ratio of 16
Since the stoichiometric ratio of [tex]\(NaOH\)[/tex] is smaller (16 < 20), [tex]\(NaOH\)[/tex] is the limiting reactant.
5. Determine the excess reactant:
- Since [tex]\(NaOH\)[/tex] is the limiting reactant, [tex]\(NBr_3\)[/tex] is the excess reactant.
6. Calculate the amount of [tex]\(NBr_3\)[/tex] remaining after the reaction:
[tex]\[
\text{Used } NaOH = 48 \text{ moles}
\][/tex]
- From the stoichiometric ratio, 2 moles of [tex]\(NBr_3\)[/tex] react with 3 moles of [tex]\(NaOH\)[/tex].
- Therefore, to react with 48 moles of [tex]\(NaOH\)[/tex], the required [tex]\(NBr_3\)[/tex] is:
[tex]\[
\left(\frac{2}{3}\right) \times 48 = 32 \text{ moles of } NBr_3
\][/tex]
- The initial amount of [tex]\(NBr_3\)[/tex] is 40 moles, so the excess [tex]\(NBr_3\)[/tex] is:
[tex]\[
40 - 32 = 8 \text{ moles}
\][/tex]
Given this analysis, the excess reactant is [tex]\(NBr_3\)[/tex], and 8 moles of [tex]\(NBr_3\)[/tex] remain unreacted. Thus, the answer to the question is:
[tex]\[ \boxed{NBr_3} \][/tex]
