Answer :
To determine the mass of iron (\( Fe \)) produced from the reaction between iron(III) oxide (\( Fe_2O_3 \)) and aluminum (\( Al \)), we can follow these steps:
1. Write the Balanced Chemical Equation:
[tex]\[ Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 \][/tex]
This tells us that one mole of \( Fe_2O_3 \) reacts with two moles of \( Al \) to produce two moles of \( Fe \) and one mole of \( Al_2O_3 \).
2. Determine the Molar Masses:
- Molar mass of \( Fe_2O_3 \) is:
[tex]\[ (2 \times 55.85 \, \text{g/mol}) + (3 \times 16.00 \, \text{g/mol}) = 159.69 \, \text{g/mol} \][/tex]
- Molar mass of \( Al \) is:
[tex]\[ 26.98 \, \text{g/mol} \][/tex]
- Molar mass of \( Fe \) is:
[tex]\[ 55.85 \, \text{g/mol} \][/tex]
3. Calculate the Moles of Each Reactant:
- Moles of \( Fe_2O_3 \):
[tex]\[ \frac{60.0 \, \text{g}}{159.69 \, \text{g/mol}} = 0.3757 \, \text{moles} \][/tex]
- Moles of \( Al \):
[tex]\[ \frac{29.0 \, \text{g}}{26.98 \, \text{g/mol}} = 1.0749 \, \text{moles} \][/tex]
4. Determine the Limiting Reactant:
According to the balanced equation, 1 mole of \( Fe_2O_3 \) reacts with 2 moles of \( Al \). Therefore, we need:
[tex]\[ \text{Moles of } Al \text{ required} = 2 \times \text{Moles of } Fe_2O_3 = 2 \times 0.3757 = 0.7514 \, \text{moles} \][/tex]
Since we have 1.0749 moles of \( Al \), which is more than 0.7514 moles, \( Fe_2O_3 \) is the limiting reactant.
5. Calculate the Moles of \( Fe \) Produced:
According to the balanced equation, 1 mole of \( Fe_2O_3 \) produces 2 moles of \( Fe \). Therefore, moles of \( Fe \) produced:
[tex]\[ 0.3757 \, \text{moles} \times 2 = 0.7514 \, \text{moles} \][/tex]
6. Calculate the Mass of \( Fe \) Produced:
[tex]\[ \text{Mass of } Fe = \text{Moles of } Fe \times \text{Molar mass of } Fe = 0.7514 \, \text{moles} \times 55.85 \, \text{g/mol} = 41.97 \, \text{g} \][/tex]
Thus, the mass of iron produced is:
[tex]\[ m_{Fe} = 41.97 \, \text{grams} \][/tex]
1. Write the Balanced Chemical Equation:
[tex]\[ Fe_2O_3 + 2Al \rightarrow 2Fe + Al_2O_3 \][/tex]
This tells us that one mole of \( Fe_2O_3 \) reacts with two moles of \( Al \) to produce two moles of \( Fe \) and one mole of \( Al_2O_3 \).
2. Determine the Molar Masses:
- Molar mass of \( Fe_2O_3 \) is:
[tex]\[ (2 \times 55.85 \, \text{g/mol}) + (3 \times 16.00 \, \text{g/mol}) = 159.69 \, \text{g/mol} \][/tex]
- Molar mass of \( Al \) is:
[tex]\[ 26.98 \, \text{g/mol} \][/tex]
- Molar mass of \( Fe \) is:
[tex]\[ 55.85 \, \text{g/mol} \][/tex]
3. Calculate the Moles of Each Reactant:
- Moles of \( Fe_2O_3 \):
[tex]\[ \frac{60.0 \, \text{g}}{159.69 \, \text{g/mol}} = 0.3757 \, \text{moles} \][/tex]
- Moles of \( Al \):
[tex]\[ \frac{29.0 \, \text{g}}{26.98 \, \text{g/mol}} = 1.0749 \, \text{moles} \][/tex]
4. Determine the Limiting Reactant:
According to the balanced equation, 1 mole of \( Fe_2O_3 \) reacts with 2 moles of \( Al \). Therefore, we need:
[tex]\[ \text{Moles of } Al \text{ required} = 2 \times \text{Moles of } Fe_2O_3 = 2 \times 0.3757 = 0.7514 \, \text{moles} \][/tex]
Since we have 1.0749 moles of \( Al \), which is more than 0.7514 moles, \( Fe_2O_3 \) is the limiting reactant.
5. Calculate the Moles of \( Fe \) Produced:
According to the balanced equation, 1 mole of \( Fe_2O_3 \) produces 2 moles of \( Fe \). Therefore, moles of \( Fe \) produced:
[tex]\[ 0.3757 \, \text{moles} \times 2 = 0.7514 \, \text{moles} \][/tex]
6. Calculate the Mass of \( Fe \) Produced:
[tex]\[ \text{Mass of } Fe = \text{Moles of } Fe \times \text{Molar mass of } Fe = 0.7514 \, \text{moles} \times 55.85 \, \text{g/mol} = 41.97 \, \text{g} \][/tex]
Thus, the mass of iron produced is:
[tex]\[ m_{Fe} = 41.97 \, \text{grams} \][/tex]
