[jennaw23]

Calculate the enthalpy change;

3FeO(s) + 2CO (g)  3Fe (s) + 3CO2  (g)


Rx 1:  3Fe2O3 (s) + 3CO (g)  2Fe3O4 (s) + CO2 (g)    ∆H = -47.0 kJ

Rx 2:  Fe2O3 (s) + 3 CO (g)  2Fe (s) + 3CO2 (g)    ∆H = -25.0 kJ

Rx 3:  Fe3O4 (s) + CO (g)  3FeO (s) + CO2 (g)      ∆H = +19.0 kJ


I have tried numerous times to get this to work, but can't for the life of me figure it out. It's my belief that I have to flip/3 the third reaction and divide the second reaction by 2

[chenbeier]

The first equation is wrong.

Rx 1:  3Fe2O3 (s) + 3CO (g)  2Fe3O4 (s) + CO2 (g)    ∆H = -47.0 kJ

has to be Rx 1:  3Fe2O3 (s) + CO (g)  2Fe3O4 (s) + CO2 (g)    ∆H = -47.0 kJ

I got + 14 kJ

[jennaw23]

Sorry, the teacher made a mistake, the final should be;
3FeO(s) + 2CO (g)  3Fe (s) + 3CO2  (g)

[chenbeier]

It is still wrong

on left side you have 2 CO and on right side 3 CO₂