Question
When iron reacts with oxygen, it forms iron oxide, or rust.
4 Fe + 3 O2 2 Fe2O3
If 112 g of iron (Fe) combines with 24 g of oxygen (O2), how much iron oxide is formed?
4 Fe + 3 O2 2 Fe2O3
If 112 g of iron (Fe) combines with 24 g of oxygen (O2), how much iron oxide is formed?
Answers
GPT-4o mini
To determine how much iron oxide (Fe₂O₃) is formed from the reaction of iron and oxygen, we can first analyze the balanced chemical equation:
\[
4 \, \text{Fe} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Fe}_2\text{O}_3
\]
From this reaction, we see that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron oxide.
**Step 1: Calculate the number of moles of Fe and O₂.**
First, we need the molar masses:
- Molar mass of iron (Fe) = 55.85 g/mol
- Molar mass of oxygen (O₂) = 32.00 g/mol
Now we calculate the moles of iron and oxygen provided:
- Moles of Fe = \(\frac{112 \, \text{g}}{55.85 \, \text{g/mol}} \approx 2.00 \, \text{mol}\)
- Moles of O₂ = \(\frac{24 \, \text{g}}{32.00 \, \text{g/mol}} = 0.75 \, \text{mol}\)
**Step 2: Determine the limiting reactant.**
According to the stoichiometry of the equation:
- 4 moles of Fe react with 3 moles of O₂.
We need to find how many moles of O₂ are required for the available moles of Fe:
- Moles of O₂ required for 2.00 moles of Fe = \(\frac{3 \, \text{mol O}_2}{4 \, \text{mol Fe}} \times 2.00 \, \text{mol Fe} = 1.50 \, \text{mol O}_2\)
Since we only have 0.75 moles of O₂, O₂ is the limiting reactant.
**Step 3: Calculate how much Fe₂O₃ can be produced.**
According to the stoichiometry of the balanced equation:
- 3 moles of O₂ produce 2 moles of Fe₂O₃.
- We can find how many moles of Fe₂O₃ can be produced from 0.75 moles of O₂:
\[
\text{Moles of Fe}_2\text{O}_3 = \frac{2 \, \text{mol Fe}_2\text{O}_3}{3 \, \text{mol O}_2} \times 0.75 \, \text{mol O}_2 = 0.50 \, \text{mol Fe}_2\text{O}_3
\]
**Step 4: Convert moles of Fe₂O₃ to grams.**
Now we can calculate the mass of Fe₂O₃ formed. The molar mass of Fe₂O₃ is:
- Molar mass of Fe₂O₃ = \( (2 \times 55.85) + (3 \times 16.00) = 159.7 \, \text{g/mol}\)
Now we convert the moles of Fe₂O₃ to grams:
\[
\text{Mass of Fe}_2\text{O}_3 = 0.50 \, \text{mol} \times 159.7 \, \text{g/mol} = 79.85 \, \text{g}
\]
**Conclusion:**
Therefore, the mass of iron oxide (Fe₂O₃) formed from the reaction is approximately **79.85 g**.
\[
4 \, \text{Fe} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Fe}_2\text{O}_3
\]
From this reaction, we see that 4 moles of iron react with 3 moles of oxygen to produce 2 moles of iron oxide.
**Step 1: Calculate the number of moles of Fe and O₂.**
First, we need the molar masses:
- Molar mass of iron (Fe) = 55.85 g/mol
- Molar mass of oxygen (O₂) = 32.00 g/mol
Now we calculate the moles of iron and oxygen provided:
- Moles of Fe = \(\frac{112 \, \text{g}}{55.85 \, \text{g/mol}} \approx 2.00 \, \text{mol}\)
- Moles of O₂ = \(\frac{24 \, \text{g}}{32.00 \, \text{g/mol}} = 0.75 \, \text{mol}\)
**Step 2: Determine the limiting reactant.**
According to the stoichiometry of the equation:
- 4 moles of Fe react with 3 moles of O₂.
We need to find how many moles of O₂ are required for the available moles of Fe:
- Moles of O₂ required for 2.00 moles of Fe = \(\frac{3 \, \text{mol O}_2}{4 \, \text{mol Fe}} \times 2.00 \, \text{mol Fe} = 1.50 \, \text{mol O}_2\)
Since we only have 0.75 moles of O₂, O₂ is the limiting reactant.
**Step 3: Calculate how much Fe₂O₃ can be produced.**
According to the stoichiometry of the balanced equation:
- 3 moles of O₂ produce 2 moles of Fe₂O₃.
- We can find how many moles of Fe₂O₃ can be produced from 0.75 moles of O₂:
\[
\text{Moles of Fe}_2\text{O}_3 = \frac{2 \, \text{mol Fe}_2\text{O}_3}{3 \, \text{mol O}_2} \times 0.75 \, \text{mol O}_2 = 0.50 \, \text{mol Fe}_2\text{O}_3
\]
**Step 4: Convert moles of Fe₂O₃ to grams.**
Now we can calculate the mass of Fe₂O₃ formed. The molar mass of Fe₂O₃ is:
- Molar mass of Fe₂O₃ = \( (2 \times 55.85) + (3 \times 16.00) = 159.7 \, \text{g/mol}\)
Now we convert the moles of Fe₂O₃ to grams:
\[
\text{Mass of Fe}_2\text{O}_3 = 0.50 \, \text{mol} \times 159.7 \, \text{g/mol} = 79.85 \, \text{g}
\]
**Conclusion:**
Therefore, the mass of iron oxide (Fe₂O₃) formed from the reaction is approximately **79.85 g**.