Calculating the Amount of Iron(III) Oxide Required for Iron Production

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Iron(III) oxide (Fe2O3), also known as hematite, is a key oxidant in the steel-making process and plays a significant role in iron production. By understanding and applying stoichiometry, we can determine the exact amount of iron(III) oxide needed to produce a specific amount of iron. This article will walk through the calculation steps required for this process.

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Stoichiometry and Chemical Reaction

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The balanced chemical equation for the reduction of iron(III) oxide to iron is as follows:

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Fe2O3(s) 3C(s) → 2Fe(l) 3CO(g)

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From the equation, we can see that 1 mole of Fe2O3 produces 2 moles of Fe. This relationship is the cornerstone for our calculation.

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Step-by-Step Calculation

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To find out how many grams of Fe2O3 are required to produce 4.65 grams of Fe, we will proceed with the following steps:

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Step 1: Calculate Moles of Fe Produced

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The first step is to calculate the moles of Fe produced from the given mass of Fe. The molar mass of Fe (iron) is approximately 55.85 g/mol.

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Moles of Fe  frac{mass of Fe}{molar mass of Fe}  frac{4.65 , text{g}}{55.85 , text{g/mol}} approx 0.0832 , text{mol}

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Step 2: Calculate Moles of Fe2O3 Required

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From the balanced equation, 2 moles of Fe are produced from 1 mole of Fe2O3. Therefore, the moles of Fe2O3 required can be calculated as:

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Moles of Fe2O3  frac{Moles of Fe}{2}  frac{0.0832 , text{mol}}{2} approx 0.0416 , text{mol}

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Step 3: Calculate Grams of Fe2O3 Required

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To find the mass of Fe2O3 required, we need to calculate its molar mass:

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Molar mass of Fe2O3  2 times 55.85   3 times 16.00  111.7   48.00  159.7 , text{g/mol}

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Now we can find the mass of Fe2O3 required:

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Mass of Fe2O3  Moles of Fe2O3 times Molar mass of Fe2O3  0.0416 , text{mol} times 159.7 , text{g/mol} approx 6.64 , text{g}

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Conclusion

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Therefore, approximately 6.64 grams of Fe2O3 are required to produce 4.65 grams of Fe.

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Alternatively, using a similar method, the calculation can be simplified for the balanced equation:

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2 Fe2O3 → 4 Fe 3 O2

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After reaction, 4.65 grams of Fe is produced. Change it into moles:

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No of moles  frac{mass in grams}{atomic mass}  frac{4.65 , text{g}}{55.8 , text{g/mol}} approx 0.083 , text{mol}

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Equation shows that Fe2O3 produces double moles of Fe after reaction. And the number of moles of Fe2O3 are half than Fe:

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No of moles of Fe  0.083 , text{mol}No of moles of Fe2O3  frac{0.083}{2} approx 0.041 , text{mol}

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Calculate the mass in grams of Fe2O3 using the number of moles:

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No of moles  frac{mass in grams}{formula mass}Formula mass of Fe2O3  159.3 , text{g/mol}0.041  frac{mass in grams}{159.3}mass in grams  0.041 times 159.3 approx 6.53 , text{g}

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Hence, approximately 6.53 grams of Fe2O3 are required to produce 4.65 grams of Fe.