Certainly! Let's go through the detailed, step-by-step process of determining whether aluminum (Al) is the limiting reactant and find the mass of the excess Fe2O3 remaining.
### Step-by-Step Solution:
#### 1. Determine Molar Masses
First, we need the molar masses of the reactants:
- Molar mass of Fe2O3 (iron(III) oxide): [tex]\( 159.69 \, \text{g/mol} \)[/tex]
- Molar mass of Al (aluminum): [tex]\( 26.98 \, \text{g/mol} \)[/tex]
#### 2. Calculate Moles of Each Reactant
Next, we convert the given masses of Fe2O3 and Al to moles:
- Moles of Fe2O3:
[tex]\[
\text{Moles of Fe2O3} = \frac{\text{mass of Fe2O3}}{\text{molar mass of Fe2O3}} = \frac{201 \, \text{g}}{159.69 \, \text{g/mol}} \approx 1.259 \, \text{mol}
\][/tex]
- Moles of Al:
[tex]\[
\text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{64.7 \, \text{g}}{26.98 \, \text{g/mol}} \approx 2.398 \, \text{mol}
\][/tex]
#### 3. Balanced Chemical Equation and Molar Ratio
The balanced chemical equation for the reaction is:
[tex]\[
\text{Fe2O3} + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al2O3}
\][/tex]
From this equation, we see that:
- 1 mole of Fe2O3 reacts with 2 moles of Al.
#### 4. Determine the Limiting Reactant
We compare the moles ratio to decide which reactant limits the reaction:
- For 1.259 moles of Fe2O3: we would need [tex]\( 1.259 \times 2 = 2.518 \)[/tex] moles of Al.
- We have only 2.398 moles of Al available.
Since we need 2.518 moles of Al but only have 2.398 moles, aluminum (Al) is the limiting reactant.
#### 5. Calculate the Moles of Fe2O3 that Reacted
Now we find out how much Fe2O3 actually reacts with the available aluminum:
- Moles of Fe2O3 that react
[tex]\[
\text{Moles of Fe2O3 reacted} = \frac{\text{moles of Al}}{2} = \frac{2.398}{2} = 1.199 \, \text{moles}
\][/tex]
#### 6. Calculate the Excess Fe2O3 Remaining
- Moles of Fe2O3 remaining (excess) are:
[tex]\[
\text{Moles of Fe2O3 excess} = \text{initial moles of Fe2O3} - \text{moles of Fe2O3 reacted} = 1.259 \, \text{mol} - 1.199 \, \text{mol} = 0.059652 \, \text{mol}
\][/tex]
- Convert this back to grams using the molar mass of Fe2O3:
[tex]\[
\text{Mass of excess Fe2O3} = \text{moles of Fe2O3 excess} \times \text{molar mass of Fe2O3} = 0.059652 \, \text{mol} \times 159.69 \, \text{g/mol} \approx 9.53 \, \text{g}
\][/tex]
### Conclusion
The excess Fe2O3 remaining after the reaction is approximately 9.53 grams to three significant figures.
