Sure, let's go through the steps required to determine the percent yield of NH₃ in this reaction.
### 1. Finding the Moles of H₂:
First, we need to determine how many moles of H₂ we have.
The molar mass of H₂ is 2.016 g/mol.
Given:
- Mass of H₂ = 26.3 g
Using the formula:
[tex]\[ \text{Moles of } H_2 = \frac{\text{mass}}{\text{molar mass}} \][/tex]
[tex]\[ \text{Moles of } H_2 = \frac{26.3 \, \text{g}}{2.016 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of } H_2 \approx 13.04563 \, \text{mol} \][/tex]
### 2. Using the Balanced Equation:
The balanced chemical equation indicates that 3 moles of H₂ produce 2 moles of NH₃.
[tex]\[ N_2 + 3H_2 \rightarrow 2NH_3 \][/tex]
So, the moles of NH₃ produced can be calculated by using the stoichiometry of the reaction.
From the balanced equation, we know:
[tex]\[ \text{Moles of } NH_3 = \left( \frac{2}{3} \right) \text{Moles of } H_2 \][/tex]
[tex]\[ \text{Moles of } NH_3 = \left( \frac{2}{3} \right) \times 13.04563 \][/tex]
[tex]\[ \text{Moles of } NH_3 \approx 8.69709 \, \text{mol} \][/tex]
### 3. Calculating Theoretical Yield of NH₃:
The molar mass of NH₃ is 17.03052 g/mol.
So, the theoretical yield of NH₃ in grams is:
[tex]\[ \text{Theoretical yield of } NH_3 = \text{moles of } NH_3 \times \text{molar mass of } NH_3 \][/tex]
[tex]\[ \text{Theoretical yield of } NH_3 = 8.69709 \, \text{mol} \times 17.03052 \, \text{g/mol} \][/tex]
[tex]\[ \text{Theoretical yield of } NH_3 \approx 148.11596 \, \text{g} \][/tex]
### 4. Calculating Percent Yield:
The actual yield given is 79.0 g.
Percent yield is calculated using the formula:
[tex]\[ \% \text{ Yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \][/tex]
[tex]\[ \% \text{ Yield} = \left( \frac{79.0 \, \text{g}}{148.11596 \, \text{g}} \right) \times 100 \][/tex]
[tex]\[ \% \text{ Yield} \approx 53.34\% \][/tex]
So, the percent yield of NH₃ is approximately [tex]\(53.4 \%\)[/tex].
