Solving Iron Metal Production in a Blast Furnace

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SUMMARY

The production of iron metal in a blast furnace involves the reduction of iron(III) oxide (Fe2O3) with carbon monoxide (CO), represented by the balanced equation: Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g). To calculate the enthalpy change (∆H) for this reaction, participants discussed three relevant equations, including their respective enthalpy values. The challenge lies in balancing the equations correctly to avoid fractions, particularly with Fe3O4, which complicates the calculation. The consensus emphasizes the importance of using whole numbers in stoichiometric coefficients for clarity.

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  • Understanding of chemical reactions and stoichiometry
  • Familiarity with enthalpy changes and thermodynamic principles
  • Knowledge of the balanced equations for iron oxide reduction
  • Ability to manipulate chemical equations for calculations
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  • Study the principles of thermodynamics related to chemical reactions
  • Learn how to balance complex chemical equations effectively
  • Explore the concept of enthalpy and its calculation in chemical processes
  • Investigate the industrial applications of blast furnaces in metallurgy
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Chemistry students, chemical engineers, and professionals in metallurgy who are involved in the production and optimization of iron metal through blast furnace processes.

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I have a problem I can't seem to solve. here it is and here is what I have gotten.

Iron metal is produced in a blast furnace through a complex series of reactions the involve reduction of iron(III) oxide with carbon monoxide

a. Write a balanced overall equation for the process, including the other product.


Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

b. Use the equations below the calculate ∆H of rxn for the overall equation:

3Fe2O3(s) + CO(g) → Fe3O4(s) + CO2(g) ∆Ho = -48.5 kJ

Fe(s) + CO2(g) → FeO(s) + CO(g) ∆Ho = -11.0 kJ

Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ∆Ho = 22.0 kJ


This what I come up with, i get 2/3 Fe3O4 on one side and 1/3 Fe3O4 on the other side giving me a 1/3 of Fe3O4 that dose not cancel each other out.

Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ∆H= ?
Fe2O3(s) + 1/3CO(g) → 1/3Fe3O4(s) + 1/3CO2(g) ∆Ho = -16.17kJ
2FeO(s) + 2CO(g) → 2 Fe(s) + 2CO2(g) ∆Ho = +22.0 kJ
2/3Fe3O4(s) + 2/3CO(g) → 2FeO(s) + 2/3CO2(g) ∆Ho = +14.67 kJ
 
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devon10 said:
I have a problem I can't seem to solve. here it is and here is what I have gotten.

Iron metal is produced in a blast furnace through a complex series of reactions the involve reduction of iron(III) oxide with carbon monoxide

a. Write a balanced overall equation for the process, including the other product.


Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

b. Use the equations below the calculate ∆H of rxn for the overall equation:

3Fe2O3(s) + CO(g) → Fe3O4(s) + CO2(g) ∆Ho = -48.5 kJ

Fe(s) + CO2(g) → FeO(s) + CO(g) ∆Ho = -11.0 kJ

Fe3O4(s) + CO(g) → 3FeO(s) + CO2(g) ∆Ho = 22.0 kJ


This what I come up with, i get 2/3 Fe3O4 on one side and 1/3 Fe3O4 on the other side giving me a 1/3 of Fe3O4 that dose not cancel each other out.

Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ∆H= ?
Fe2O3(s) + 1/3CO(g) → 1/3Fe3O4(s) + 1/3CO2(g) ∆Ho = -16.17kJ
2FeO(s) + 2CO(g) → 2 Fe(s) + 2CO2(g) ∆Ho = +22.0 kJ
2/3Fe3O4(s) + 2/3CO(g) → 2FeO(s) + 2/3CO2(g) ∆Ho = +14.67 kJ

You don't really need 4 equations, just stick with those you got. Try again using the three equations given to you. When you have done it is like you say Fe3O4 is not going to cancel each other out but remember that is ok to multiply a equation with a number. It is easier when you avoid fractions.
 

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