Solving Iron Metal Production in a Blast Furnace

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Iron metal production in a blast furnace involves the reduction of iron(III) oxide with carbon monoxide, represented by the balanced equation Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g). The discussion focuses on calculating the enthalpy change (∆H) for this reaction using provided equations, but a participant encounters difficulty in balancing the equations, particularly with Fe3O4. Suggestions include simplifying the approach by avoiding fractions and multiplying equations for clarity. The importance of correctly applying thermodynamic principles to derive the overall reaction enthalpy is emphasized. Accurate calculations are essential for understanding the blast furnace process.
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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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