Answers by visitors named: A Canadian

Whoops, never mind, I got the answer.

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'...

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...

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 "?

But I don't think division would mean "which" weighs 18g..

, but then would 1mol/18g H2O mean 1 mol per 18 grams per mol of H2O..? There's one mol for every 18g of H2O..? It's confusing

I just realised I made the stupidest mistake ever. I guess this is what happens when you try to do chemistry while sick. The last bit of math doesn't make sense; it should be 3.82 = (460.24)(Tf-21.4°C) 3.82 = Tf460.24 - 9849.136 -Tf460.24 = -9849.136 - 3....

I checked it over for what feels like the tenth time and I see that I should've converted the 3.82 kJ to 3820 J; I ended up with the correct answer using that. But I feel like my method was a bit over-complicated. Is there a simpler method that I'm just o...

If you absolutely had to improve it what would you improve? Anything that sticks out especially to you But thank you; haha, I wouldn't go after an American prize anyways!

Sorry, how would you solve for mols? I feel like this should be obvious, but ...