chemistry
posted by Andy on .
Can you please check if I came up with the right answer for this please.
A 4.30L container had 55.10 moles of N2O4 added to it. When equilibrium was established the concentration of N2O4 was found to be 1.69M.What is the Kc for the following reaction?
N2O4(g)<>2NO2(g)
Reasoning and workings:
N2O4<>2No2
12.814M of O
12.814x2x
12.814x=1.69
x=11.124M
Kc=[NO2]^2/[N2o4]
=22.248^2/1.69
=292.88M
Hows that ??
Is it right??
If it is i am going to be so happy!!
Thanks andy

I have (N2O4) = 55.10/4.30 = 12.01 M BUT you ended up with the right value for x and the right value for NO2 as well as the right value for K.

Doess this mean I fluked the write answer??
Could you write it out as I have above please with your logical answers so I can go away and study if we are right..I am never comfortable not knowing I have the right workings
Thanks andy 
No, you were right. I read my calculator wrong. My eyesight leaves something to be desired and I read 12.81 as 12.01. So all of your numbers are correct. See below for the only picky thing I could find.
N2O4<>2No2
12.814M of O This should be 12.814 M N2O4, not of O. Everything else looks ok to me.
12.814x2x
12.814x=1.69
x=11.124M
Kc=[NO2]^2/[N2o4]
=22.248^2/1.69
=292.88M 
Thanks DR BOB,
You had me worried for a second there!!
Thanks for all your great help on this one
Regards Andy