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
- 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
- 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
- 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
- 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
- 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.