Post a New Question


posted by .

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.......)??

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question