Question
How many grams of iron can be obtained from 265 grams of Fe₃O₄
Answers
GPT-4o mini
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.
1. **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:
\[
\text{Molar mass of Fe}_3\text{O}_4 = (3 \times 55.85 \, \text{g/mol}) + (4 \times 16.00 \, \text{g/mol})
= 167.55 \, \text{g/mol} + 64.00 \, \text{g/mol}
= 231.55 \, \text{g/mol}
\]
2. **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:
\[
\text{Mass of Fe in one mole of Fe}_3\text{O}_4 = 3 \times 55.85 \, \text{g} = 167.55 \, \text{g}
\]
3. **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.
1. **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:
\[
\text{Molar mass of Fe}_3\text{O}_4 = (3 \times 55.85 \, \text{g/mol}) + (4 \times 16.00 \, \text{g/mol})
= 167.55 \, \text{g/mol} + 64.00 \, \text{g/mol}
= 231.55 \, \text{g/mol}
\]
2. **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:
\[
\text{Mass of Fe in one mole of Fe}_3\text{O}_4 = 3 \times 55.85 \, \text{g} = 167.55 \, \text{g}
\]
3. **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.