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Chemistry question
If someone could please help me. I’m not sure how to do this and I’m stressing out over it.
If 3.25 L of NH3 react with an excess of oxygen at 500 degrees C and 5.00 atm pressure, what volume of water will be produced at the same temperature and pressure?
4NH3(g) + 5O2(g)—> 6H2O(g) + 4NO(g)
A.) 2.04 L
B.) 2.17 L
C.) 3.33 L
D.) 4.14 L
E.) 4.88 L
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sorry poster, I don’t have any chemistry textbooks to look this up. Can I borrow yours?
Anonymous edited this post 1 year ago. Read the previous text »
Chemistry question
If 3.25 L of NH3 react with an excess of oxygen at 500 degrees C and 5.00 atm pressure, what volume of water will be produced at the same temperature and pressure?
4NH3(g) + 5O2(g)—> 6H2O(g) + 4NO(g)
A.) 2.04 L
B.) 2.17 L
C.) 3.33 L
D.) 4.14 L
E.) 4.88 L
ok, that’s better… give me a little while and I’ll see what I can do for you
I have notes and stuff but we just haven’t gone over how to find oxygen when you are starting out with another substance. I’ve tried several different ways and I’m not even coming close. So I don’t know what to do. I would really appreciate your help though. :)
ok, here’s a site on balancing chemical equations.
it looks like an easy to follow page, so give it a try and let me know if it helps.
it’s been far too long since I was last in a chemistry class, but I am pretty good at finding solutions online!
well we don’t have to balance the equation it’s finding the volume of water that is produced. So I’m not sure if I have to use the equation for a conversion or what.
as a matter of interest, what state’s chemistry curriculum is this? I might be able to search it that way
here’s a site with links.
you go there and check some out while I check others..
Thank you I did end up finding out the answer. Thank you for using your time to help me out. It was greatly appreciated!
good, what was the answer? lol
thanks. now I have to figure it out or it’ll drive me crazy!!
lol
have fun
I’ll give you some clues. You have to find moles and then end up using the equation M1V1=M2V2
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