To determine how many grams of iron can be obtained from 3.0 kg of iron(II,III) oxide (Fe₃O₄), we first need to perform some calculations based on the molar masses of the compounds involved.
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Calculating the molar mass of Fe₃O₄:
- The atomic mass of iron (Fe) is approximately 55.85 g/mol.
- The molar mass of Fe₃O₄ can be calculated as follows: \[ \text{Molar mass of Fe}_3\text{O}_4 = (3 \times 55.85) + (4 \times 16.00) = 167.55 + 64.00 = 231.55 \text{ g/mol} \]
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Converting mass of Fe₃O₄ to grams: Since we have 3.0 kg of Fe₃O₄, we need to convert this mass into grams: \[ 3.0 \text{ kg} = 3000 \text{ g} \]
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Calculating the moles of Fe₃O₄: Now we can find the number of moles of Fe₃O₄: \[ \text{Moles of Fe}_3\text{O}_4 = \frac{\text{mass of Fe}_3\text{O}_4}{\text{molar mass of Fe}_3\text{O}_4} = \frac{3000 \text{ g}}{231.55 \text{ g/mol}} \approx 12.95 \text{ mol} \]
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Using the stoichiometry of the reaction: From the formula Fe₃O₄, we see that each mole of Fe₃O₄ contains 3 moles of Fe. Therefore: \[ \text{Moles of Fe} = 3 \times \text{Moles of Fe}_3\text{O}_4 = 3 \times 12.95 \text{ mol} \approx 38.85 \text{ mol} \]
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Calculating the mass of iron obtained: Now we can convert moles of iron to grams: \[ \text{Mass of Fe} = \text{Moles of Fe} \times \text{molar mass of Fe} = 38.85 \text{ mol} \times 55.85 \text{ g/mol} \approx 2165.85 \text{ g} \]
Therefore, the mass of iron that can be obtained from 3.0 kg of Fe₃O₄ is approximately 2165.85 grams of iron.