Equilibrium Concentrations?? please show work so i can compare answers.?

Question

At 25 degrees C, the equilibrium constant Kc for the reaction is 0.035. A mixture of 8.00 moles B and 12.00 moles of C in a 20.0L container is allowed to come into equilibrium. What is the equilibrium concentration of A?
2A (g) yields B(g)+C(g)

thanks for the help, it makes alot more sense now, I had it set up right, but i kept getting messed up solving for x with the quadratic, thanks

Answer 1

Initial concentration B: 8/20=0.4 M
Initial concentration C: 12/20= 0.6 M

.. .. .. .. .. .. 2A <=> B + C
Initial .. .. .. .. .. .. .. 0.4 .. 0.6
React .. .. .. .. .. .. .. x .. ..x
Produce .. 2x
At Equil. .. . 2x .. ..0.4-x.. 0.6-x

K=[B][C]/[A]^2 =>
0.035=(0.4-x)(0.6-x)/(2x)^2 =>
0.035*4x^2= 0.24 -0.4x -0.6x + x^2 =>
0.86x^2 -x+0.24 =0
It is a quadratic equation of the type ax^2+bx+c=0
There are two possible solutions:

x1= (-b+Squareroot( b^2-4ac))/2a
x2= (-b-Squareroot( b^2-4ac))/2a

x1 =(-(-1)+ SQRT( (-1)^2 -4*0.86*0.24) ) / (2*0.86)= 0.824
x2=0.339

x1 is rejected since it is bigger than the concentration of B which is the limiting reagent (and also bigger of C) and we can't have negative values for the equilibrium concentration of B (and C)
Thus the only acceptable solution is x=x2= 0.339
[A]=2*x= 2*0.339= 0.678 M

Answer 2

You should express B and C in concentration.
[B] = 8/20 = 0.40M
[C] = 12/20 = 0.60M

Inital concentrations:
[A]=0 [B] = 0.40M [C] = 0.60M
Change in concentration:
A: +2x B: -x C: -x
Final Concentration:
[A] = 2x [B] = 0.40 - x [C] = 0.60 - x

K = (0.4-x)(0.6-x)/(2x)^2
0.035 = (0.4-x)(0.6-x)/4x^2

using the equilibrium expression K = [products]/[reactants].

0.035 * 4x^2 = (0.4-x)(0.6-x)
0.14 x^2 = 0.24 - x + x^2
0.86x^2 - 0.1x + 0.24 = 0

The solutions to this quadratic equation are 0.824 and 0.339. 0.824 is not a valid solution, as [B] and [C] are less than 0.824 (mathematically, this would give negative equilibrium concentrations). So x = 0.339 is the correct solution.

[A] = 2x = 0.678M
[B] = 0.4 - x = 0.061M
[C] = 0.6 - x = 0.261M

If the algebra in setting up the quadratic equation was correct, then we should be able to put these values into the equilibrium expression and get 0.035

K = 0.061 * 0.261 / (0.678)^2 = 0.0346.