To determine how many grams of iron can be obtained from 265 grams of Fe₃O₄ (magnetite), we need to use the molar mass of Fe₃O₄ and the stoichiometry of the compound.
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Calculate the molar mass of Fe₃O₄:
- Iron (Fe) has a molar mass of approximately 55.85 g/mol.
- Oxygen (O) has a molar mass of approximately 16.00 g/mol.
Therefore, for Fe₃O₄: \[ \text{Molar mass of Fe}_3\text{O}_4 = (3 \times 55.85 , \text{g/mol}) + (4 \times 16.00 , \text{g/mol}) = 167.55 , \text{g/mol} + 64.00 , \text{g/mol} = 231.55 , \text{g/mol} \]
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Calculate the mass of iron in Fe₃O₄:
Since there are 3 moles of iron in each mole of Fe₃O₄, the mass of iron can be calculated as follows: \[ \text{Mass of Fe in one mole of Fe}_3\text{O}_4 = 3 \times 55.85 , \text{g} = 167.55 , \text{g} \] -
Determine the amount of iron in 265 grams of Fe₃O₄:
We will use the ratio of the mass of iron to the mass of Fe₃O₄ to find out how much iron is in 265 grams of Fe₃O₄: \[ \text{Mass of Fe} = \left( \frac{167.55 , \text{g Fe}}{231.55 , \text{g Fe}_3\text{O}_4} \right) \times 265 , \text{g Fe}_3\text{O}_4 \]Performing the calculation: \[ \text{Mass of Fe} = \frac{167.55}{231.55} \times 265 \approx 237.32 , \text{g Fe} \]
Therefore, from 265 grams of Fe₃O₄, approximately 237.32 grams of iron can be obtained.