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Chemistry Help

I’m having a few problems with a Chemistry practice test, any input or tips or anything at all would be helpful.

I have this reaction that I have balanced: Fe3O4 + 4H2 > 3Fe + 4H2O

And I have to answer the following questions:

a. How many grams of Fe3O4 are needed to produce 500 grams of Fe.
b. How many moles of H2O are produced when 25 kg of H2 react.
c. How many milligrams of Fe are produced when 10 grams of Fe3O4 and 2 grams of H2 are reacted.
d. How much of what reactant is left over in part c above.
e. If the Fe3O4 is present in an ore containing 5% Fe3O4 and 95% inert material, how much ore is required to produce 150 grams of H2O?
f. if 11.26 kilograms of Fe3O4 and 1.26 kilograms of H2 are mixed, how many moles of Fe is formed and which reactant is in excess?

I will be working on these later and I haven’t really been able to understand the way the book describes them so any help is appreciated.

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.

Those questions look hard…

so you joined a bodybuilding forum to get help with your chemistry homework…

e. If the Fe3O4 is present in an ore containing 5% Fe3O4 and 95% inert material, how much ore is required to produce 150 grams of H2O?

Simply find the amount of Fe3O4 needed to produce 150 grams of H2O (find number of moles of H2O in 150 g, this is 4x the number of moles of Fe3O4 you need so divide by four, then convert back to grams of Fe3O4 by using molar mass), then divide by 0.05 (or multiply by 20) to find the amount of ore needed)

f. if 11.26 kilograms of Fe3O4 and 1.26 kilograms of H2 are mixed, how many moles of Fe is formed and which reactant is in excess?

Do the same thing you did for c to find limiting reactant (except in this case the process would be grams to moles of Fe3O4 and H2, then there need to be over 4x as many moles of H2 as Fe3O4 for Fe3O4 to be the limiting, otherwise the H2 will be limiting. Whichever is limiting, find the number of moles of product one mole of that substance will make (3/4 for H2 and 3 for Fe3O4) and multiply by the number of moles you start with. This will be the amount of product formed)

That should be right, sorry for the chickenshit presentation.

[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

[quote]ramrod63 wrote:
so you joined a bodybuilding forum to get help with your chemistry homework…[/quote]

That’s what I was wondering. Maybe he thinks since some bodybuilders juice, they have an in-depth understanding of chemistry…maybe.

Nice avatar btw.

Chemistry rocked. Unfortunatly I forgot it all after pre-med.

You can’t take a few minutes to essentially plug some values into a couple of equations?

chemistry was awesome, but, alas, i have forgotten it all.

eyes glaze over instantly.

I hated Gen Chem so much. If you like chemistry but don’t like algebra then Organic will be a blast for you.
Oh, and x2 on what pzehtoeur said.

Hey, I’m in grade 11 chemistry right now!
Woohoo!

And I got a question exactly like that, you in my class db?
(It’s a dead give away that I’m a fat asian guy).

This is just basic stoichiometry and limiting reagent shit. Just convert your shit to mole:mole ratios, then convert it to the units you need. If you can’t handle this, I don’t see you passing.

I hated that shit in college. If you want my advice go find a guy named Barry. He’s been a grad student for like 30 years and smells really bad. But he’ll tutor you for free and he’s smart. Just dont accept his invitations to go camping.

Whoa, that got a little too personal for a minute. That’s what I did in chem 101 and 102. I got a D in 102 so dont take my advice.

[quote]db7890 wrote:
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.
[/quote]

So you want the answers actually given to you? Either do the work yourself or fail the course.