Sure, let's determine the percent yield of the reaction step-by-step. The reaction given is:
[tex]\[ \text{Fe}_2\text{O}_3(s) + 2 \text{Al}(s) \rightarrow \text{Al}_2\text{O}_3(s) + 2 \text{Fe}(s) \][/tex]
We are given:
- Actual yield of Fe: [tex]\( 28.65 \, \text{g} \)[/tex]
- Initial mass of Fe₂O₃: [tex]\( 50.00 \, \text{g} \)[/tex]
- Molar mass of Fe₂O₃: [tex]\( 159.7 \, \text{g/mol} \)[/tex]
- Molar mass of Fe: [tex]\( 55.85 \, \text{g/mol} \)[/tex]
### Step 1: Calculate the moles of Fe₂O₃
To find the moles of Fe₂O₃, use its initial mass and molar mass:
[tex]\[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass of Fe}_2\text{O}_3}{\text{Molar mass of Fe}_2\text{O}_3} = \frac{50.00 \, \text{g}}{159.7 \, \text{g/mol}} = 0.313087 \, \text{mol} \][/tex]
### Step 2: Determine the moles of Fe produced
From the stoichiometry of the reaction, 1 mole of Fe₂O₃ produces 2 moles of Fe. Using the moles of Fe₂O₃ calculated:
[tex]\[ \text{Moles of Fe} = \text{Moles of Fe}_2\text{O}_3 \times 2 = 0.313087 \, \text{mol} \times 2 = 0.626174 \, \text{mol} \][/tex]
### Step 3: Calculate the theoretical yield of Fe
The theoretical yield is found by converting the moles of Fe to grams using its molar mass:
[tex]\[ \text{Theoretical yield of Fe} = \text{Moles of Fe} \times \text{Molar mass of Fe} = 0.626174 \, \text{mol} \times 55.85 \, \text{g/mol} = 34.971822 \, \text{g} \][/tex]
### Step 4: Calculate the percent yield
Finally, we determine the percent yield using the actual yield and theoretical yield:
[tex]\[ \text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 = \left( \frac{28.65 \, \text{g}}{34.971822 \, \text{g}} \right) \times 100 = 81.923 \% \][/tex]
Thus, the percent yield of the reaction is approximately [tex]\( 81.923 \% \)[/tex].
