4Fe + 3O2 -> 2Fe2O3 + 1.65E3 kJ
C) what is the enthalpy change for the formation of 23.6 g of iron(iii) oxide?
I'm trying to understand thermochemical reactions and stoichiometry, so could you please tell me if this is correct? And please explain the parts that I clearly don't understand:
The question, is it the same thing as asking, "how much energy was released if iron(iii) oxide had a mass of 23.6 g?"
According to the equation:
4 moles of Fe and 3 moles of O2 react to release -1.65E3 kJ. ....then 2 moles of Fe "contains" -1.65E3 kJ??
To calculate how much kJ of energy was released when the mass of Fe2O3 was 23.6g, we need the amount of moles:
23.6 g are in 1 mole of Fe2O3, and every mole of Fe2O3 is 160g, the molar mass. (??) which, I think??, explains why this works :
23.6g Fe2O3 x 1 mol Fe2O3/160g
And Since there's -1.65E3 kJ per 2 moles of Fe2O3: -1.65E3/2 mol Fe2O3
Putting it altogether:
Enthalpy = 23.6 g Fe2O3 x 1 mol Fe2O3/160g x -1.65E3kJ/2 mol 2Fe2O3
= 1.22E2 kJ
I'm hoping I've got the units all right too...
Also, what does it mean if "there's -1.65E3 kJ per 2 moles of Fe2O3"? Then there's -1.65E3 kJ 'contained' in the Fe2O3??? Because we wouldn't say -1.65E3 kJ is RELEASED when Fe2O3 reacts, because Fe2O3 is the product... So what exactly is going on (between Fe2O3 and the energy or enthalpy change or whatever.......)??
Chemistry - DrBob222, Wednesday, February 6, 2013 at 8:57pm
I saw your post and the same kind of questions with Bob Pursley. I suppose you're still struggling. I think you are making this a lot harder than it is. Your 122E3 kJ is correct and the equation you used looks ok to me. Here is how I do them
1.65E3 kJ energy is released when 2*160g (that's 320 g) Fe2O3 is formed. How much is released to form 23.6g Fe2O3?. That's
1.65E3 kJ x (23.6g/320g) = 122 kJ.