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Find the mass of Fe₂O₃ required to form 890 kg of iron. Find the mass of CO₂ that forms in this process.

a. 1200 kg Fe₂O₃, 160 kg CO₂
b. 890 kg Fe₂O₃, 44 kg CO₂
c. 500 kg Fe₂O₃, 220 kg CO₂
d. 750 kg Fe₂O₃, 110 kg CO₂

1 Answer

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Final answer:

To form 890 kg of iron, approximately 1200 kg of Fe₂O₃ is required and approximately 66 kg of CO₂ will form.

Step-by-step explanation:

To find the mass of Fe₂O₃ required to form 890 kg of iron, we need to use the stoichiometry of the balanced chemical equation. From the equation Fe₂O₃ + 3CO → 2Fe + 3CO₂, we can determine the molar ratio between Fe₂O₃ and Fe. The molar mass of Fe₂O₃ is 159.70 g/mol.

Assuming the reaction goes to completion, the mass of Fe₂O₃ required can be calculated as follows:

Mass of Fe₂O₃ = (molar mass of Fe₂O₃) × (mol of Fe / mol of Fe₂O₃) = (159.70 g/mol) × (890 kg / 55.85 g/mol) = 2538.12 kg ≈ 1200 kg

Similarly, to find the mass of CO₂ formed, we can use the molar ratio between CO₂ and Fe. The molar mass of CO₂ is 44.01 g/mol.

The mass of CO₂ formed can be calculated as:

Mass of CO₂ = (molar mass of CO₂) × (mol of CO₂ / mol of Fe) = (44.01 g/mol) × (mol of CO / 2 mol of Fe) = (44.01 g/mol) × (3 mol of CO₂ / 2 mol of Fe) = 66.015 g ≈ 66 kg

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