Question

A three-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?

Answer 1

From 390 kg of Fe2O3, approximately 143.51 kg of iron metal would be generated after applying the given yields for each step of the reduction process.

Step-by-step explanation:

The task here is to calculate the mass of iron metal that can be produced from the reduction of ferric oxide (Fe2O3) through a series of steps with specified yields. We're given the initial mass of Fe2O3 and the yields of each step to work with:

1. 3Fe2O3 + CO → 2Fe3O4 + CO2 (84.6% yield)
2. Fe3O4 + CO → 3FeO + CO2 (50.8% yield)
3. FeO + CO → Fe + CO2 (85.7% yield)

For a 390 kg sample of Fe2O3, here's how we proceed:

• First step yield: 390 kg × 84.6% = 329.94 kg of Fe3O4
• Second step yield: 329.94 kg × 50.8% = 167.41 kg of FeO
• Third step yield: 167.41 kg × 85.7% = 143.51 kg of Fe

Therefore, from 390 kg of Fe2O3, approximately 143.51 kg of iron metal would be produced, considering the given yields for each reaction step and that carbon monoxide (CO) is in excess.

Answer 2

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

Step-by-step 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