Question

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

Answer 1

To determine how many kilograms of magnetite (Fe₃O₄) are needed to obtain 185 kg of pure iron, we first need to understand the stoichiometry of the reaction.

Fe₃O₄ contains iron (Fe) and oxygen (O) in the following ratio:

- 3 moles of Fe in 1 mole of Fe₃O₄.

### Step 1: Molar Mass Calculation
1. Calculate the molar mass of Fe₃O₄:
- Molar mass of Fe = 55.85 g/mol
- Molar mass of O = 16.00 g/mol

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

### Step 2: Iron Contribution
2. Calculate how much iron is contained in 1 mole of Fe₃O₄:
- 3 moles of Fe = 3 × 55.85 g = 167.55 g of Fe.

### Step 3: Finding Moles of Iron
3. Determine how many moles of Fe correspond to 185 kg (185,000 g) of pure iron:
\[
\text{Moles of Fe} = \frac{185,000 \text{ g}}{55.85 \text{ g/mol}} \approx 3307.92 \text{ moles}
\]

### Step 4: Moles of Fe₃O₄ Needed
4. Since 3 moles of Fe come from 1 mole of Fe₃O₄, calculate the moles of Fe₃O₄ needed:
\[
\text{Moles of Fe}_3\text{O}_4 = \frac{3307.92 \text{ moles of Fe}}{3} \approx 1102.64 \text{ moles of Fe}_3\text{O}_4
\]

### Step 5: Mass of Fe₃O₄
5. Now calculate the mass of Fe₃O₄ needed:
\[
\text{Mass of Fe}_3\text{O}_4 = 1102.64 \text{ moles} \times 231.55 \text{ g/mol} \approx 255,000 \text{ g} = 255 \text{ kg}
\]

### Conclusion
To obtain 185 kg of pure iron, approximately **255 kg** of Fe₃O₄ are needed.