Athree-step process for producing molten iron metal from fe2o3 is: 3fe2o3 + co → 2fe3o4 + co2 fe3o4 + co → 3feo + co2 feo + co → fe + co2 assuming that the reactant co is present in excess and that the yields, respectively, for the three steps are 84.6%, 50.8% and 85.7%, what mass of iron metal would be produced from 390. kg of fe2o3?

There is 100.4652 kg of iron produced from 390 kg of Fe2O3

Explanation:

Step 1: Given data

3Fe2O3 + CO → 2Fe3O4 + CO2

Fe3O4 + CO → 3FeO + CO2

FeO + CO → Fe + CO2

Mass of Fe2O3 = 390 kg = 390000 grams

Molar mass of Fe2O3 = 159.69 g/moles

Step 2: Calculate moles of Fe2O3

Moles of Fe2O3 = mass of Fe2O3 / Molar mass of Fe2O3

Moles of Fe2O3 = 390000 grams / 159.69 g/moles = 2442.2 moles

Step 3: Calculate expected moles of Fe3O4

In the first equation, for 3 moles of Fe2O3 consumed ,we get 2 moles of Fe3O4. The mole ratio is 3:2

This means if we consume 2442.2 moles of Fe2O3, there will be produced 2/3 * 2442.2 = 1628.2 moles of Fe3O4

Since the yield for the step is only 84.6 %

This will be 0.846 * 1628.2 = 1377.4 moles of Fe3O4

Step 4: Calculate expected moles of FeO

In the second equation, for 1 mole of Fe3O4 consumed, there is produced 3 moles of FeO

This means for 1377.4 moles of Fe3O4 consumed, there is 3*1377.4 = 4132.2 moles of FeO produced

Since the yield for the step is only 50.8%

This will be 0.508 * 4132.2 = 2099.2 moles of FeO

Step 5: Calculate expected moles of Fe

In the third equation, for 1 mole of FeO consumed, there is produced 1 mole of Fe.

This means for 2099.2 moles of FeO consumed, there is also 2099.2 moles of Fe produced

Since the yield is only 85.7%

This will be 0.857 * 2099.2 = 1799 moles of Fe

Step 6: Calculate mass of Fe

Mass of Fe = moles of Fe * Molar mass of Fe

Mass of Fe = 1799 moles of Fe * 55.845 g/moles = 100465.2 grams = 100.4652 kg of Fe

There is 100.4652 kg of iron produced from 390 kg of Fe2O3

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