10 mole of nitrous oxide in 2dm3 container decompose to nitrogen and oxygen at equilibrium the concentration of nitrous oxide is 2.20 mole calculate kc

Answer 1

(N2O) = 10 mols/2 dm3 = 5 Molar
..................2N2O ==> 2N2 + O2
I...................5................0........0
C..................-2x.............2x......x
E.................5-2x............2x.......x
Kc = (N2)^2(O2)/(N2O)
We have a problem here because of your post. Is that 2.20 mole in the problem 2.20 molar or is it 2.20 mole. I will assume it is 2.20 mol; however, you say that is a concentration and it can't be. If it is concentration it means 2.20 MOLAR. Again, I'll assume 2.20 mol N2O at equilibrium in the 2 L container so (N2O) at equilibrium will be 2.20/2 = 1.1 M.
So you know 5-2x = 1.1. Solve for x = 1.95 M. That means N2O is 1.1 M and O2 = x = 1.95 M and N2 is 2*1.95 = 3.9 M. Plug those values into the Kc expression bove and solve for Kc.
IF YOU MEAN that N2O at equilibrium is 2.20 molar, then
5-2x = 2.20, then x = 1.4. So in assumption, (N2O) = 2.20, (N2) = 2.8 and (O2) = 1.4. Plug those numbers into Kc expression and solve for Kc. Post your work if you get stuck. I hope this isn't confusing but you must learn to post problems correctly for it makes a difference it you the concentration is 2.20 M or if you have 2.20 mols.