To determine how many kilograms of Fe₃O₄ (magnetite) are needed to obtain 185 kg of pure iron, we first need to look at the chemical formula of Fe₃O₄ and its composition.
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Molar Mass of Fe₃O₄:
- Iron (Fe) molar mass = 55.85 g/mol
- Oxygen (O) molar mass = 16.00 g/mol
For Fe₃O₄: \[ \text{Molar mass of Fe}_3\text{O}_4 = (3 \times 55.85) + (4 \times 16.00) = 167.55 + 64.00 = 231.55 , \text{g/mol} \]
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Molar Mass Calculation:
- The total mass of iron in one mole of Fe₃O₄ is \(3 \times 55.85 , \text{g} = 167.55 , \text{g}\).
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Iron Yield:
- From 1 mole of Fe₃O₄ (231.55 g), we can obtain 167.55 g of pure iron.
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Calculating Required Fe₃O₄:
- To obtain 185 kg of pure iron, we set up a ratio: \[ \frac{167.55 , \text{g Fe}}{231.55 , \text{g Fe}_3\text{O}_4} = \frac{185000 , \text{g Fe}}{x , \text{g Fe}_3\text{O}_4} \] Where \(x\) is the mass of Fe₃O₄ required.
Rearranging gives us: \[ x , \text{g Fe}_3\text{O}_4 = \frac{185000 , \text{g Fe} \times 231.55 , \text{g Fe}_3\text{O}_4}{167.55 , \text{g Fe}} \]
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Calculating \(x\): \[ x = \frac{185000 \times 231.55}{167.55} \approx \frac{42754250}{167.55} \approx 255,399.98 , \text{g} \]
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Convert to kg: \[ x \approx 255.40 , \text{kg} \]
Thus, to obtain 185 kg of pure iron, you would need approximately 255.40 kg of Fe₃O₄.