Chemical Equation - check, please?

Question

Iron can be extraced from Iron oxide (Fe₂O₃) by chemical reduction in the presence of Carbon. In the blast furnace involving heating of carbaron, CO₂and molten Iron can be produced.

Fe₂O₃ C ---->(heat) Fe(s) CO₂(g)

Thanks for any help regarding this uncertainty.

Answer 1

if this equation happens that would be

2FeO₃  + 3C ---> 4Fe + 3CO₂

Answer 2

you have to balance it, so on the product side, there is a 2 infrom on iron, a 3 in front of carbon dioxide, and on the reactant side - a 2 in front of oxygen and a 3 in front of carbon... i think

Answer 3

For a detailed description of blast furnace operation, see:

http://www.steel.org/AM/Template.cfm?Sec...

The coke provides carbon monoxide which reacts according to the scheme below:

1) 3 Fe2O3 + CO = CO2 + 2 Fe3O4 Begins at 850° F

2) Fe3O4 + CO = CO2 + 3 FeO           Begins at 1100° F

3) FeO + CO = CO2 + Fe

    or

    FeO + C = CO + Fe

The overall reaction is Fe2O3 + 3CO → 2Fe + 3CO2

Answer 4

This process is also called smelting. Smelting is heating of an oxide mineral (in this case Fe2O3) with coke C to obtain a pure metal. If I'm not mistaken, carbon doesn't take place in the reaction at all.

Fe2O3(s) + heat <--> 2Fe(s) + 3/2O2(g)

change in Gibbs free energy reaction = 743.6kJ/mol

As you can see this reaction has a very high Gibbs free energy which means the reaction is not favored. The reason why coke is used is to lower the Gibbs free energy. This happens because the coke removes oxygen causing equilibrium to shift toward the formation of products.

C(s) + 1/2O2(g) --> CO2(g)

change in Gibbs free energy reaction = -143.8kJ/mol

So when you add the reactions together the Gibbs of the overall reaction is lower.

2Fe2O3(s) + 3C(s) + heat <--> 4Fe(s) + 3CO2(g)

change in Gibbs free energy overall = 599.8kJ/mol

The overall Gibbs free energy is significantly reduced which mean a tremendously less amount of energy is used after coke is added.

Tips: Don't forget the sign conventions like (s), (g), and (l) the phases are very important. You did this correctly, but when you read a problem make sure your molecular formula is the right one. This is the most important make sure the equation is balanced with the correct coeficcents.  

I hope this wasn't confusing.