To determine which reaction mixture would produce the greatest amount of ammonia ([tex]\(NH_3\)[/tex]), we need to compare the amounts of the product formed from different mixtures. The balanced chemical equation provided is:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
This indicates that 1 mole of nitrogen gas ([tex]\(N_2\)[/tex]) reacts with 3 moles of hydrogen gas ([tex]\(H_2\)[/tex]) to produce 2 moles of ammonia ([tex]\(NH_3\)[/tex]).
Let's analyze each mixture given:
1. Mixture 1:
- [tex]\(N_2\)[/tex]: 2 moles
- [tex]\(H_2\)[/tex]: 6 moles
2. Mixture 2:
- [tex]\(N_2\)[/tex]: 1 mole
- [tex]\(H_2\)[/tex]: 8 moles
3. Mixture 3:
- [tex]\(N_2\)[/tex]: 5 moles
- [tex]\(H_2\)[/tex]: 15 moles
4. Mixture 4:
- [tex]\(N_2\)[/tex]: 3 moles
- [tex]\(H_2\)[/tex]: 9 moles
For each mixture, the amount of [tex]\(NH_3\)[/tex] produced is determined by the limiting reactant, i.e., the reactant that will be completely consumed first in the reaction. Let's calculate the limiting reactant and the amount of [tex]\(NH_3\)[/tex] produced for each mixture:
1. Mixture 1 Analysis:
- [tex]\(N_2\)[/tex]: 2 moles can produce [tex]\(2 \times 2 = 4\)[/tex] moles of [tex]\(NH_3\)[/tex].
- [tex]\(H_2\)[/tex]: 6 moles can produce [tex]\(2 \times \frac{6}{3} = 4\)[/tex] moles of [tex]\(NH_3\)[/tex].
Both reactants produce 4 moles of [tex]\(NH_3\)[/tex]. The limiting reactant is not clear, but the result is 4 moles of [tex]\(NH_3\)[/tex].
2. Mixture 2 Analysis:
- [tex]\(N_2\)[/tex]: 1 mole can produce [tex]\(2 \times 1 = 2\)[/tex] moles of [tex]\(NH_3\)[/tex].
- [tex]\(H_2\)[/tex]: 8 moles can produce [tex]\(2 \times \frac{8}{3} \approx 5.33\)[/tex] moles of [tex]\(NH_3\)[/tex].
The limiting reactant is [tex]\(N_2\)[/tex], producing 2 moles of [tex]\(NH_3\)[/tex].
3. Mixture 3 Analysis:
- [tex]\(N_2\)[/tex]: 5 moles can produce [tex]\(2 \times 5 = 10\)[/tex] moles of [tex]\(NH_3\)[/tex].
- [tex]\(H_2\)[/tex]: 15 moles can produce [tex]\(2 \times \frac{15}{3} = 10\)[/tex] moles of [tex]\(NH_3\)[/tex].
Both reactants produce 10 moles of [tex]\(NH_3\)[/tex]. The limiting reactant is not clear, but the result is 10 moles of [tex]\(NH_3\)[/tex].
4. Mixture 4 Analysis:
- [tex]\(N_2\)[/tex]: 3 moles can produce [tex]\(2 \times 3 = 6\)[/tex] moles of [tex]\(NH_3\)[/tex].
- [tex]\(H_2\)[/tex]: 9 moles can produce [tex]\(2 \times \frac{9}{3} = 6\)[/tex] moles of [tex]\(NH_3\)[/tex].
Both reactants produce 6 moles of [tex]\(NH_3\)[/tex]. The limiting reactant is not clear, but the result is 6 moles of [tex]\(NH_3\)[/tex].
Now let's summarize the results:
- Mixture 1 produces 4 moles of [tex]\(NH_3\)[/tex].
- Mixture 2 produces 2 moles of [tex]\(NH_3\)[/tex].
- Mixture 3 produces 10 moles of [tex]\(NH_3\)[/tex].
- Mixture 4 produces 6 moles of [tex]\(NH_3\)[/tex].
Therefore, the mixture that produces the greatest amount of ammonia is:
Mixture 3 with 10 moles of [tex]\(NH_3\)[/tex].
