Question
A 2,3 kg sample of iron oxide, Fe_2O_3, is added to 1,7 kg of Carbon monoxide, according to the following balanced equation : 3Fe_2O_3 (s) +CO(g) ->2Fe_3O_4(s)+CO_2 (g)
Identity the limiting reagent in this reaction by means of a suitable calculation
Identity the limiting reagent in this reaction by means of a suitable calculation
Answers
GPT-4o mini
To determine the limiting reagent in the reaction between iron oxide (Fe₂O₃) and carbon monoxide (CO), we need to follow these steps:
1. **Find the molar masses of the reactants**:
- For Fe₂O₃:
- Iron (Fe): 55.85 g/mol, Oxygen (O): 16.00 g/mol
- Molar mass of Fe₂O₃ = (2 × 55.85) + (3 × 16.00) = 111.7 + 48.00 = 159.7 g/mol
- For CO:
- Carbon (C): 12.01 g/mol, Oxygen (O): 16.00 g/mol
- Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol
2. **Convert mass of reactants to moles**:
- For Fe₂O₃:
\[
\text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2300 \text{ g}}{159.7 \text{ g/mol}} \approx 14.39 \text{ mol}
\]
- For CO:
\[
\text{Moles of CO} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1700 \text{ g}}{28.01 \text{ g/mol}} \approx 60.7 \text{ mol}
\]
3. **Determine the stoichiometry of the reaction**:
The balanced equation is:
\[
3\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{Fe}_3\text{O}_4 + \text{CO}_2
\]
From the equation, we see that 3 moles of Fe₂O₃ react with 1 mole of CO. Therefore, the mole ratio is:
\[
\frac{\text{Moles of Fe}_2\text{O}_3}{\text{Moles of CO}} = \frac{3}{1}
\]
4. **Calculate required moles of CO for the available moles of Fe₂O₃**:
Using the moles of Fe₂O₃ we have (14.39 mol), we determine how much CO is required:
\[
\text{Moles of CO required} = \frac{1 \text{ mol CO}}{3 \text{ mol Fe}_2\text{O}_3} \times 14.39 \text{ mol Fe}_2\text{O}_3 = \frac{14.39}{3} \approx 4.80 \text{ mol CO}
\]
5. **Compare with available moles of CO**:
We have approximately 60.7 moles of CO available, which is more than the required 4.80 moles.
6. **Identify the limiting reagent**:
Since we need only about 4.80 moles of CO to completely react with 14.39 moles of Fe₂O₃, and we have more than sufficient CO, the limiting reagent in this reaction is **Fe₂O₃**.
1. **Find the molar masses of the reactants**:
- For Fe₂O₃:
- Iron (Fe): 55.85 g/mol, Oxygen (O): 16.00 g/mol
- Molar mass of Fe₂O₃ = (2 × 55.85) + (3 × 16.00) = 111.7 + 48.00 = 159.7 g/mol
- For CO:
- Carbon (C): 12.01 g/mol, Oxygen (O): 16.00 g/mol
- Molar mass of CO = 12.01 + 16.00 = 28.01 g/mol
2. **Convert mass of reactants to moles**:
- For Fe₂O₃:
\[
\text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2300 \text{ g}}{159.7 \text{ g/mol}} \approx 14.39 \text{ mol}
\]
- For CO:
\[
\text{Moles of CO} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1700 \text{ g}}{28.01 \text{ g/mol}} \approx 60.7 \text{ mol}
\]
3. **Determine the stoichiometry of the reaction**:
The balanced equation is:
\[
3\text{Fe}_2\text{O}_3 + \text{CO} \rightarrow 2\text{Fe}_3\text{O}_4 + \text{CO}_2
\]
From the equation, we see that 3 moles of Fe₂O₃ react with 1 mole of CO. Therefore, the mole ratio is:
\[
\frac{\text{Moles of Fe}_2\text{O}_3}{\text{Moles of CO}} = \frac{3}{1}
\]
4. **Calculate required moles of CO for the available moles of Fe₂O₃**:
Using the moles of Fe₂O₃ we have (14.39 mol), we determine how much CO is required:
\[
\text{Moles of CO required} = \frac{1 \text{ mol CO}}{3 \text{ mol Fe}_2\text{O}_3} \times 14.39 \text{ mol Fe}_2\text{O}_3 = \frac{14.39}{3} \approx 4.80 \text{ mol CO}
\]
5. **Compare with available moles of CO**:
We have approximately 60.7 moles of CO available, which is more than the required 4.80 moles.
6. **Identify the limiting reagent**:
Since we need only about 4.80 moles of CO to completely react with 14.39 moles of Fe₂O₃, and we have more than sufficient CO, the limiting reagent in this reaction is **Fe₂O₃**.