[H Mac]

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     Determine the quantity, in moles, of the gas that remains unreacted.
« on: January 30, 2008, 05:16:39 PM » Quote Modify  

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Consider the following reation, which takes place in an autoclave at 250'C and 800 atm.
       NH3(g)  + 7/4 O2(g) -->  NO2 (g) +  3/2 H2O(g)
Into the reaction vessel has been placed 200L of NH3(g) and 120L of O2(g). The reaction is allowed to go to completion.
Determine the quantity, in moles, of the gas that remains unreacted.


Just some hints as to what I might do here would be fantastic!
  
I know I already posted it but I still havn't been able to figure it out on my own... this is what I have got though... and just looking to you guys to see if I am right!
Thank you kindly,

NH3(g) + 7/4O2(g) --> NO2(g) + 3/2H2O(g)

NH3                             O2
m = 200L                      m= 120L
M=1404.03L/mol            M=
n=  m
     M
  = 0.142 mol * 7/4
  = 0.249              therefore O2 n= 0.249

0.249 - 0.142 =  0.107 mol

Therefore there is 0.107 mol of unreacted O2 left in this reaction.

Let me know what you think of this (I had no examples to go on so please be critical)  

[Borek]

I think you were told to calculate moles first. And you really should add your post to the previous thread to not clutter the forums.