To determine the limiting reagent in the reaction between iron oxide (Fe₂O₃) and carbon monoxide (CO), we need to follow these steps:
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Find the molar masses of the reactants:
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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
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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
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Convert mass of reactants to moles:
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For Fe₂O₃: \[ \text{Moles of Fe}_2\text{O}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{2300 \text{ g}}{159.7 \text{ g/mol}} \approx 14.39 \text{ mol} \]
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For CO: \[ \text{Moles of CO} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{1700 \text{ g}}{28.01 \text{ g/mol}} \approx 60.7 \text{ mol} \]
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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: \[ \frac{\text{Moles of Fe}_2\text{O}_3}{\text{Moles of CO}} = \frac{3}{1} \]
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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: \[ \text{Moles of CO required} = \frac{1 \text{ mol CO}}{3 \text{ mol Fe}_2\text{O}_3} \times 14.39 \text{ mol Fe}_2\text{O}_3 = \frac{14.39}{3} \approx 4.80 \text{ mol CO} \]
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Compare with available moles of CO: We have approximately 60.7 moles of CO available, which is more than the required 4.80 moles.
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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.
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