A three-step process for producing molten iron metal from Fe2O3 is:3Fe2O3 + CO → 2Fe3O4 + CO2Fe3O4 + CO → 3FeO + CO2FeO + CO → Fe + CO2Assuming that the reactant CO is present in excess and that the yields, respectively, for the three steps are 80.1%, 67.6% and 75.0%, what mass of iron metal would be produced from 390. kg of Fe2O3?

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Asked Oct 3, 2019
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A three-step process for producing molten iron metal from Fe2O3 is:

  1. 3Fe2O3 + CO → 2Fe3O4 + CO2
  2. Fe3O4 + CO → 3FeO + CO2
  3. FeO + CO → Fe + CO2

Assuming that the reactant CO is present in excess and that the yields, respectively, for the three steps are 80.1%, 67.6% and 75.0%, what mass of iron metal would be produced from 390. kg of Fe2O3?

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Expert Answer

Step 1

The moles of iron produced from 390.kg o...

given, Fe O, 390.kg=390000 g
molar massof Fe,03=159.7 g/mol
390000g-2442.07 mol Fe,O,
159.7 g/mol
moles of Fe O
From each reaction,moles can be calculatedas,
3Fe2O+CO2Fe,O4+CO2 yield=80.1%
2mol FeO, x0.80 1=1304.06 mol FegO,
2442.07mol FeO;^ 3mol Fe,03
Fe O4+CO3FeO+CO2
yield-67.6%
3mol FeO
=1304.06 mol Fe O
-x0.676=2644.63 mol FeO
X
1mol Fe,O
yield 75.0%
FeO+COFe+CO2
2644.63mol FeOx_lmolFex0.75=1983.47 mol Fe
1mol FeO
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given, Fe O, 390.kg=390000 g molar massof Fe,03=159.7 g/mol 390000g-2442.07 mol Fe,O, 159.7 g/mol moles of Fe O From each reaction,moles can be calculatedas, 3Fe2O+CO2Fe,O4+CO2 yield=80.1% 2mol FeO, x0.80 1=1304.06 mol FegO, 2442.07mol FeO;^ 3mol Fe,03 Fe O4+CO3FeO+CO2 yield-67.6% 3mol FeO =1304.06 mol Fe O -x0.676=2644.63 mol FeO X 1mol Fe,O yield 75.0% FeO+COFe+CO2 2644.63mol FeOx_lmolFex0.75=1983.47 mol Fe 1mol FeO

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