Question

How many kg of Fe₃O₄ are needed to obtain 185 kg of pure iron?

Answer 1

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.

1. **Molar Mass of Fe₃O₄**:
- Iron (Fe) molar mass = 55.85 g/mol
- Oxygen (O) molar mass = 16.00 g/mol

For Fe₃O₄:
\[
\text{Molar mass of Fe}_3\text{O}_4 = (3 \times 55.85) + (4 \times 16.00) = 167.55 + 64.00 = 231.55 \, \text{g/mol}
\]

2. **Molar Mass Calculation**:
- The total mass of iron in one mole of Fe₃O₄ is \(3 \times 55.85 \, \text{g} = 167.55 \, \text{g}\).

3. **Iron Yield**:
- From 1 mole of Fe₃O₄ (231.55 g), we can obtain 167.55 g of pure iron.

4. **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:
\[
x \, \text{g Fe}_3\text{O}_4 = \frac{185000 \, \text{g Fe} \times 231.55 \, \text{g Fe}_3\text{O}_4}{167.55 \, \text{g Fe}}
\]

5. **Calculating \(x\)**:
\[
x = \frac{185000 \times 231.55}{167.55} \approx \frac{42754250}{167.55} \approx 255,399.98 \, \text{g}
\]

6. **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**.