[quote]pzehtoeur wrote:

I don’t have the answers but here are the steps:

a) Figure out how many moles of Fe is 500 grams. Once you know that, you can set up a proportion to figure out how many moles of Fe3O4 you need (1 mole of Fe3O4 for every 3 moles of Fe). Once you figure out how many moles of Fe3O4, convert back to grams using the molar mass.

b)Again, figure out how many moles of H2 is 25kg. Set up a proportion (4 moles of H2 for every 4 moles of H2O) and you should be able to figure it out. Since the mole ratios of H2 to H2O is 1, then once you figure out how many moles of H2 you have, it will be the moles of H2O.

c) Figure out how many moles of Fe3O4 is 10 grams and how many moles of H2 is 2 grams. You will need 4 times as many moles of H2 in order to match up with the Fe3O4 since for every mole of Fe3O4, you will require 4 moles of H2. If you have excess Fe3O4 (that is, your ratio of moles of H2 to Fe3O4 is less than 4), then your limiting reagent will be H2. If you have excess H2 (your ratio of moles of H2 to Fe3O4 is greater than 4), then your limiting reagent will be Fe3O4. Take the limiting reagent and take that number of moles and again set up a proportion: 1 mole of Fe3O4 for every 3 moles of Fe or 4 moles of H2 for every 3 moles of Fe. This will be how many moles of Fe are produced.

d) Whatever was NOT your limiting reagent in part C, take what was used up (the number of moles) and subtract that from the original number of moles. That will be how many moles will be left over. If you want to convert back to grams, divide by the molar mass.

e) and f) I actually have to run to class right now so I quickly wrote down these steps, which is why I did not solve them. If you still need help, I’ll be happy to help you although I’m sure someone else on here will have helped you solved them by the time I have returned.

I hope what I wrote above is somewhat clear. [/quote]

It looks pretty clear but I don’t know if I will have time today to actually solve them, if someone could help me out with that it’d be great.

And no I did not join this site for the sole purpose of chemistry help lol