To determine which reactant should be used to calculate the yield, we need to identify the limiting reactant in the chemical reaction. The limiting reactant is the one that will be completely consumed first, thus determining the maximum amount of product that can be formed. Here's a step-by-step solution:
1. Calculate the moles of [tex]\(H_2\)[/tex]:
- Given: mass of [tex]\(H_2\)[/tex] = 24.0 g
- Molar mass of [tex]\(H_2\)[/tex] = 2.0 g/mol
- Moles of [tex]\(H_2\)[/tex] = [tex]\(\frac{\text{mass}}{\text{molar mass}} = \frac{24.0 \text{ g}}{2.0 \text{ g/mol}} = 12.0 \text{ moles}\)[/tex]
2. Calculate the moles of [tex]\(N_2\)[/tex]:
- Given: mass of [tex]\(N_2\)[/tex] = 8.0 g
- Molar mass of [tex]\(N_2\)[/tex] = 28.0 g/mol
- Moles of [tex]\(N_2\)[/tex] = [tex]\(\frac{\text{mass}}{\text{molar mass}} = \frac{8.0 \text{ g}}{28.0 \text{ g/mol}} \approx 0.2857 \text{ moles}\)[/tex]
3. Determine the mole ratio from the balanced chemical equation:
- The balanced equation is: [tex]\(3 H_2 + N_2 \rightarrow 2 NH_3\)[/tex]
- This shows that 3 moles of [tex]\(H_2\)[/tex] react with 1 mole of [tex]\(N_2\)[/tex].
4. Calculate the actual mole ratio of [tex]\(H_2\)[/tex] to [tex]\(N_2\)[/tex]:
- Actual mole ratio = [tex]\(\frac{\text{moles of } H_2}{\text{moles of } N_2} = \frac{12.0}{0.2857} \approx 42.0\)[/tex]
5. Compare the actual mole ratio to the needed mole ratio for the reaction:
- Needed mole ratio (from the balanced equation) = 3:1 = 3
- Actual mole ratio = 42.0
6. Identify the limiting reactant:
- If the actual ratio is greater than the needed ratio (42.0 > 3), there is more [tex]\(H_2\)[/tex] in excess compared to [tex]\(N_2\)[/tex].
- This means [tex]\(N_2\)[/tex] is the limiting reactant because it will be consumed first.
7. Conclusion:
- The chemist should use nitrogen ([tex]\(N_2\)[/tex]) to calculate the yield because it is the limiting reactant.
Thus, the correct answer is:
A. Nitrogen, because the ratio of moles is greater than 3:1 for [tex]\(H_2: N_2\)[/tex].
