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Given the reaction:

[tex]\[ Fe_2O_3(s) + 2Al(s) \longrightarrow 2Fe(l) + Al_2O_3(s) \][/tex]

1. What masses of iron(III) oxide and aluminum must be used to produce 15.0 g of iron?

2. What is the maximum mass of aluminum oxide that could be produced?



Answer :

GinnyAnswer
Let's solve the problem step-by-step using the balanced chemical equation and given the desired mass of iron (Fe).

The balanced chemical equation is:
[tex]\[ \text{Fe}_2\text{O}_3(s) + 2\text{Al}(s) \longrightarrow 2\text{Fe}(l) + \text{Al}_2\text{O}_3(s) \][/tex]

1. Molar Masses:
- Molar mass of Fe (iron): [tex]\(55.845 \, \text{g/mol}\)[/tex]
- Molar mass of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] (iron(III) oxide):
[tex]\[2 \times \text{molar mass of Fe} + 3 \times \text{molar mass of O}\][/tex]
[tex]\[= 2 \times 55.845 \, \text{g/mol} + 3 \times 15.999 \, \text{g/mol}\][/tex]
[tex]\[= 111.69 \, \text{g/mol} + 47.997 \, \text{g/mol} = 159.687 \, \text{g/mol}\][/tex]
- Molar mass of Al (aluminum): [tex]\(26.982 \, \text{g/mol}\)[/tex]
- Molar mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] (aluminum oxide):
[tex]\[2 \times \text{molar mass of Al} + 3 \times \text{molar mass of O}\][/tex]
[tex]\[= 2 \times 26.982 \, \text{g/mol} + 3 \times 15.999 \, \text{g/mol}\][/tex]
[tex]\[= 53.964 \, \text{g/mol} + 47.997 \, \text{g/mol} = 101.961 \, \text{g/mol}\][/tex]

2. Calculate moles of Fe produced:
Given the mass of Fe produced is [tex]\(15.0 \, \text{g}\)[/tex],
[tex]\[ \text{moles of Fe} = \frac{\text{mass of Fe}}{\text{molar mass of Fe}} \][/tex]
[tex]\[ = \frac{15.0 \, \text{g}}{55.845 \, \text{g/mol}} \][/tex]
[tex]\[ \approx 0.2686 \, \text{moles} \][/tex]

3. Calculate moles of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] required:
According to the balanced equation, 1 mole of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] produces 2 moles of Fe. Therefore:
[tex]\[ \text{moles of Fe}_2\text{O}_3 = \frac{\text{moles of Fe}}{2} \][/tex]
[tex]\[ \approx \frac{0.2686 \, \text{moles}}{2} \][/tex]
[tex]\[ \approx 0.1343 \, \text{moles} \][/tex]

4. Calculate mass of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] required:
[tex]\[ \text{mass of Fe}_2\text{O}_3 = \text{moles of Fe}_2\text{O}_3 \times \text{molar mass of Fe}_2\text{O}_3 \][/tex]
[tex]\[ \approx 0.1343 \, \text{moles} \times 159.687 \, \text{g/mol} \][/tex]
[tex]\[ \approx 21.446 \, \text{g} \][/tex]

5. Calculate moles of Al required:
According to the balanced equation, 2 moles of Al are needed to react with 1 mole of [tex]\(\text{Fe}_2\text{O}_3\)[/tex]. Therefore:
[tex]\[ \text{moles of Al} = 2 \times \text{moles of Fe} \][/tex]
[tex]\[ \approx 2 \times 0.2686 \, \text{moles} \][/tex]
[tex]\[ \approx 0.5372 \, \text{moles} \][/tex]

6. Calculate mass of Al required:
[tex]\[ \text{mass of Al} = \text{moles of Al} \times \text{molar mass of Al} \][/tex]
[tex]\[ \approx 0.5372 \, \text{moles} \times 26.982 \, \text{g/mol} \][/tex]
[tex]\[ \approx 14.495 \, \text{g} \][/tex]

7. Calculate mass of [tex]\(\text{Al}_2\text{O}_3\)[/tex] produced:
According to the balanced equation, 1 mole of [tex]\(\text{Fe}_2\text{O}_3\)[/tex] produces 1 mole of [tex]\(\text{Al}_2\text{O}_3\)[/tex]. Therefore:
[tex]\[ \text{moles of Al}_2\text{O}_3 = \text{moles of Fe}_2\text{O}_3 \][/tex]
[tex]\[ \approx 0.1343 \, \text{moles}\][/tex]
[tex]\[ \text{mass of Al}_2\text{O}_3 = \text{moles of Al}_2\text{O}_3 \times \text{molar mass of Al}_2\text{O}_3 \][/tex]
[tex]\[ \approx 0.1343 \, \text{moles} \times 101.961 \, \text{g/mol} \][/tex]
[tex]\[ \approx 13.693 \, \text{g} \][/tex]

Summarizing:
- The mass of iron(III) oxide ([tex]\(\text{Fe}_2\text{O}_3\)[/tex]) needed: [tex]\(21.446 \, \text{g}\)[/tex]
- The mass of aluminum (Al) required: [tex]\(14.495 \, \text{g}\)[/tex]
- The maximum mass of aluminum oxide ([tex]\(\text{Al}_2\text{O}_3\)[/tex]) produced: [tex]\(13.693 \, \text{g}\)[/tex]

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