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Sure, let's go through the solution step-by-step to determine the mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] formed when 4.5 grams of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] completely reacts.
### Step 1: Determine the Molar Mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex]
First, we need to find the molar mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex].
- Iron (Fe) has an atomic mass of 55.85 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
The molar mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] can be calculated as follows:
[tex]\[ \text{Molar mass of } \text{Fe}_2 \text{O}_3 = 2 \times 55.85 + 3 \times 16.00 = 111.7 + 48.00 = 159.7 \text{ g/mol} \][/tex]
### Step 2: Calculate the Moles of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex]
Given the mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] is 4.5 grams, we can find the number of moles:
[tex]\[ \text{Moles of } \text{Fe}_2 \text{O}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5 \text{ g}}{159.7 \text{ g/mol}} \approx 0.0282 \text{ moles} \][/tex]
### Step 3: Determine the Molar Mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex]
Next, we find the molar mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex].
- Aluminum (Al) has an atomic mass of 26.98 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
The molar mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] can be calculated as follows:
[tex]\[ \text{Molar mass of } \text{Al}_2 \text{O}_3 = 2 \times 26.98 + 3 \times 16.00 = 53.96 + 48.00 = 101.96 \text{ g/mol} \][/tex]
### Step 4: Calculate the Moles and Mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] Formed
From the balanced chemical equation:
[tex]\[ \text{Fe}_2 \text{O}_3 + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al}_2 \text{O}_3 \][/tex]
The molar ratio of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] to [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] is 1:1.
Therefore, the moles of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] produced will be the same as the moles of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] reacted:
[tex]\[ \text{Moles of } \text{Al}_2 \text{O}_3 = 0.0282 \text{ moles} \][/tex]
Finally, we find the mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] formed:
[tex]\[ \text{Mass of } \text{Al}_2 \text{O}_3 = \text{moles} \times \text{molar mass} = 0.0282 \text{ moles} \times 101.96 \text{ g/mol} \approx 2.87 \text{ g} \][/tex]
### Conclusion
Therefore, the mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] formed when 4.5 grams of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] completely reacts is approximately [tex]\( \boxed{2.9 \text{ g}} \)[/tex], rounded to two significant figures.
### Step 1: Determine the Molar Mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex]
First, we need to find the molar mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex].
- Iron (Fe) has an atomic mass of 55.85 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
The molar mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] can be calculated as follows:
[tex]\[ \text{Molar mass of } \text{Fe}_2 \text{O}_3 = 2 \times 55.85 + 3 \times 16.00 = 111.7 + 48.00 = 159.7 \text{ g/mol} \][/tex]
### Step 2: Calculate the Moles of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex]
Given the mass of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] is 4.5 grams, we can find the number of moles:
[tex]\[ \text{Moles of } \text{Fe}_2 \text{O}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{4.5 \text{ g}}{159.7 \text{ g/mol}} \approx 0.0282 \text{ moles} \][/tex]
### Step 3: Determine the Molar Mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex]
Next, we find the molar mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex].
- Aluminum (Al) has an atomic mass of 26.98 g/mol.
- Oxygen (O) has an atomic mass of 16.00 g/mol.
The molar mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] can be calculated as follows:
[tex]\[ \text{Molar mass of } \text{Al}_2 \text{O}_3 = 2 \times 26.98 + 3 \times 16.00 = 53.96 + 48.00 = 101.96 \text{ g/mol} \][/tex]
### Step 4: Calculate the Moles and Mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] Formed
From the balanced chemical equation:
[tex]\[ \text{Fe}_2 \text{O}_3 + 2\text{Al} \rightarrow 2\text{Fe} + \text{Al}_2 \text{O}_3 \][/tex]
The molar ratio of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] to [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] is 1:1.
Therefore, the moles of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] produced will be the same as the moles of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] reacted:
[tex]\[ \text{Moles of } \text{Al}_2 \text{O}_3 = 0.0282 \text{ moles} \][/tex]
Finally, we find the mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] formed:
[tex]\[ \text{Mass of } \text{Al}_2 \text{O}_3 = \text{moles} \times \text{molar mass} = 0.0282 \text{ moles} \times 101.96 \text{ g/mol} \approx 2.87 \text{ g} \][/tex]
### Conclusion
Therefore, the mass of [tex]\( \text{Al}_2 \text{O}_3 \)[/tex] formed when 4.5 grams of [tex]\( \text{Fe}_2 \text{O}_3 \)[/tex] completely reacts is approximately [tex]\( \boxed{2.9 \text{ g}} \)[/tex], rounded to two significant figures.
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