For the reaction: N2+3H2<--->2NH3

Question

4.000 mol N2, 2.000 mol H2, and 6.000 mol NH3 are placed in a 2.000 L container and allowed to reach equilibrium. When equilibrium is established the [H2]=2.77 M. Calculate the equilibrium concentrations of nitrogen and ammonia. Calculate K.

Answer 1

(N2) = 4mols/2L = 2M
(H2) = 2/2 = 1M
(NH3) = 6/2 = 3M
You should add all of the zeros on the numbers below.

.......N2 + 3H2 ==> 2NH3
I.....2.0...1.0......3.0
C.....+x....+3x......-2x
E.....2.0+x.1.0+3x..3.0-2x
3.0M-2x = 2.77M
3.0-2.77 = 2x and x = 0.115
Evaluate equilibrium (N2) and (H2) and substitute into Keq expression and solve for K.
Post your work if you get stuck.
By the way, your post below with the A, B, and 2C is not quite right. That problem is done the same way as this one.