To determine how many grams of iron can be obtained from 265 grams of magnetite (Fe₃O₄), we first need to establish the molar mass of Fe₃O₄ and the amount of iron contained in it.
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Molar Mass Calculation of Fe₃O₄:
- Iron (Fe): approximately 55.85 g/mol
- Oxygen (O): approximately 16.00 g/mol
The molar mass of Fe₃O₄ can be calculated as: \[ \text{Molar mass of } \text{Fe}_3\text{O}_4 = (3 \times 55.85 \text{ g/mol}) + (4 \times 16.00 \text{ g/mol}) \] \[ = 167.55 \text{ g/mol} + 64.00 \text{ g/mol} = 231.55 \text{ g/mol} \]
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Mass of Iron in Fe₃O₄: In Fe₃O₄, there are 3 moles of iron for every mole of Fe₃O₄. The total mass of iron in one mole of Fe₃O₄ can be calculated as: \[ \text{Mass of Fe in } \text{Fe}_3\text{O}_4 = 3 \times 55.85 \text{ g} = 167.55 \text{ g} \]
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Mass Fraction of Iron: The mass fraction of iron in Fe₃O₄ can be calculated as: \[ \text{Mass fraction of Fe} = \frac{\text{Mass of Fe in Fe}_3\text{O}_4}{\text{Molar mass of Fe}_3\text{O}_4} = \frac{167.55 \text{ g}}{231.55 \text{ g}} \approx 0.723 \]
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Calculating the Mass of Iron in 265 g of Fe₃O₄: Now, we can calculate the mass of iron that can be obtained from 265 grams of Fe₃O₄: \[ \text{Mass of Fe} = \text{Mass fraction of Fe} \times \text{mass of Fe}_3\text{O}_4 \] \[ = 0.723 \times 265 \text{ g} \approx 191.595 \text{ g} \]
Thus, approximately 191.6 grams of iron can be obtained from 265 grams of Fe₃O₄.