Equilibrium Constant Calculation?

Question

Ammonium carbamate (NH2COONH4) (MM=78 g/mol) is found in the blood and urine of mammals. At a temperature of 504 K, Kc = 0.027 for the following equilibrium:
NH2COONH4(s) <=> 2 NH3(g) + CO2(g).
If 10.2 g of ammonium carbamate is introduced into a 0.500 L container, what is the total pressure inside the container at equilibrium?

Correct answer: 23.46

How would I go about doing this?

I'm pretty sure I'm doing it right, but I keep getting the wrong answer.

The equation to solve x, that I get, is 4x^3 + .027X=.00705
.......

Answer 1

The equilibrium constant: Kc = [NH3]^2[CO2]

so 0.027 = (2x)^2(x) = 4x^3

x = 0.189M = [CO2]

[NH3] = 2*[CO2], so total concentration is 2x + x = 3x = 3*0.189M = 0.567M

Use PV=nRT --> rearrange to get P = (n/V)RT, where n/V is total concentration = 0.567M. R = 0.08206 L.atm/mol.K; T = 504K. Multiply through to get 23.46 atm.

Answer 2

Since the reactant is a solid, it doesn't show up in the Keq equation. So Keq = [NH3]^2 x [CO2]
The stoichiometry predicts for every CO2 molecule, there will be 2 NH3 molecules. So [NH3] = 2x[CO2]
Substitute in and remember to square the 2. That gives you the concentration of carbon dioxide and from that you get the concentration of ammonia. Then use the general gas law PV = nRT and sub in the number of moles per liter for n/V, and you should get the pressure.