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Author: Subject: theoretical yeild
123chem
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sad.gif posted on 29-6-2004 at 10:29
theoretical yeild


anyone know how to figure this out:

N2 + 3 H2 = 2 NH3

if 1200 g of N2 is reacted with excess hydrogen, what is the theoretical yeild of NH3
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chemoleo
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[*] posted on 29-6-2004 at 10:49


Now, there is a Beginnings section to ask questions like that.

A simple solution would be (but surely not appreciated by any examiner):

1200g x 17/14 = 1457.1 g of NH3...

Don't ask what I am doing, thats for you to figure ;)




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[*] posted on 29-6-2004 at 11:05


lol.....ive got a better answer than that.....

the theoretical yield is 100 % :D
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The_Davster
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[*] posted on 29-6-2004 at 11:15


theres also this way.
1200g/28.02 x 2 x 17.04
=1459.5g of NH3

Which is about the same as Chemoleo's way, but mine is more accurate ;)
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[*] posted on 29-6-2004 at 11:17


Lets do the math:



Edit:
Someone else beat me to writing it out, so I drew it out for you.

[Edited on 6/29/2004 by sanity gone]

[Edited on 6/29/2004 by sanity gone]
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The_Davster
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[*] posted on 29-6-2004 at 11:20


sanity gone: It was suposed to be 2 mol NH3 over 1 mol N2.
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[*] posted on 29-6-2004 at 11:21


haha, I saw that right after I posted it, damn ADD.
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