At 25°c, the equilibrium constant kc for the reaction 2a(g) b(g) + c(g) is 0.035. a mixture of 8.00 moles of b and 12.00 moles of c in a 20.0 l container is allowed to come to equilibrium. what is the equilibrium concentration of a?

The chemical reaction is represented as:

2A(g) = B(g) + C(g)

To determine the equilibrium concentration of A, we make use of the equilibrium constant, Kc, given above. It is expressed as the ratio of the equilibrium concentrations of the products and the reactants. For this reaction, it is expressed as:

Kc = [B] [C] / [A]^2

From the problem statement, we are given the following 
Kc =  0.035
Volume = 20.0 L
Initial concentrations: [B] = 8.00 mol / 20.0 L = 0.4 M
                                   [C] = 12.00 mol / 20.0 L = 0.6 M
Since the initial reactants are B and C, the reaction is reversed as well as the Kc.

Kc = [A]^2 / [B][C] 

We use the ICE table:
           B              C               A
I         0.4           0.6              0
C         -x             -x             +x

E     0.4 - x     0.6 - x            x 

Kc = x^2 / (0.4-x) (0.6-x)  = 0.035

solve for x,

x = 0.07691 = [A]

Any photo? or something