yes, for each reaction in the balance equation, you get that energy.
Now for 23.6g FeOxide, you do not have two moles, you have to figure what part of the energy you get.
Enthalpy= 23.6g Fe2O3/(160g) * 1.65E3Kj/2molesFe2O3
examine the last term, it is different from yours. Why? Look at the balanced equation
This isn't exactly a homework question, I'm just trying to understand the concept we learned today...
So say we had:
4Fe + 3O2 → 2Fe2O3 + 1.65 x 10^3 kJ
Does that mean for every 4 mol of Fe, 3 mol of O2, and 2 mol of Fe2O3 we have 1.65 x 10^3 kJ?
And if I were to calculate what the enthalpy change for the formation of 23.6 g of iron(iii) oxide was,
Would it be 23.6g Fe2O3 or 23.6g of 2Fe2O3 that I would use in the equation...? When do you account for the fact that there's 2 moles of Fe2O3?
My confused attempt at solving:
Enthalpy=23.6g Fe2O3 x 1 mol Fe2O3/[(2x56) + (3x16)]g x -1.65x10^3 kJ/1 mol Fe2O3
= -243.38 kJ
Yeah, I'm pretty confused; can someone please clear things up for me??
Also, I guess it's not that important but how come when doing 1mol of Fe2O3/molar mass of Fe2O3 you can just do 1mol/160g instead of 1mol/160g/mol?? Because for molar mass we always use the unit g/mol...
Thanks so much in advance !! It's only the second lesson in grade 12 chemistry and I'm already lost! Not good.. :(
8 answers
Okay so after some trial and error (I just kept trying different things Until I got the answer), I figured out it's *probably* solved like this:
Enthalpy = 23.6g Fe2O3 x 1 mol Fe2O3/[(2x56) + (3x16)]g x -1.65x10^3 kJ/2 mol Fe2O3
= -1.22 x 10^2 kJ
If that's right, my question is, how come you divide -1.65 x 10^3 kJ by *2* mol Fe2O3 but you don't use that when you do the other calculation (I know I'm being vague - sorry; I just don't even know what the "other calculation" is even doing/calculating so I can be more specific...): *1* mol Fe2O3/[(2x56) + (3x16)]g
Here, you don't account for the fact that there's 2 mol of Fe2O3 in the equation? Why?
Basically... What's going on???
Enthalpy = 23.6g Fe2O3 x 1 mol Fe2O3/[(2x56) + (3x16)]g x -1.65x10^3 kJ/2 mol Fe2O3
= -1.22 x 10^2 kJ
If that's right, my question is, how come you divide -1.65 x 10^3 kJ by *2* mol Fe2O3 but you don't use that when you do the other calculation (I know I'm being vague - sorry; I just don't even know what the "other calculation" is even doing/calculating so I can be more specific...): *1* mol Fe2O3/[(2x56) + (3x16)]g
Here, you don't account for the fact that there's 2 mol of Fe2O3 in the equation? Why?
Basically... What's going on???
Sorry, I posted the above before I got to see your answer, bobpursley. I think I get it a little more but could you clarify a bit more based on what I asked above?
Thanks for answering!!
"Now for 23.6g FeOxide, you do not have two moles, you have to figure what part of the energy you get."
Could you clarify what you mean by that? I think this is closer to answering and fixing my confusion; what do you mean by "what part of the energy you get"?
"Now for 23.6g FeOxide, you do not have two moles, you have to figure what part of the energy you get."
Could you clarify what you mean by that? I think this is closer to answering and fixing my confusion; what do you mean by "what part of the energy you get"?
You got 1.65E3Kj for each two moles of product. You did not have two moles, you had 23.6 grams, and the point of the conversion was to figure out what part of two moles you had, then multply that by 1.65E3 Kj
I'm really sorry, but I still don't really understand - what do you mean by "figure out what part of two moles you had "?
well i think it will be 183g
what type of reaction is Mg+2HCl-->MgCl2+H2
6 answers