Question about the Ammonium carbamate?

Question

Ammonium carbamate N2H6CO2, a salt found in the blood and urine of of mammals, sublimes completely at 70oC. It dissociates into NH3 and CO2 according to the equilibrium process below.
N2H6CO2(s) « CO2(g) + 2NH3(g)

At lower temperatures, it dissociates only partially. A sample of N2H6CO2 is heated to 30.0oC and allowed to reach equilibrium. An examination of the system reveals that the total pressure of the gases is 0.155 atm. Calculate the Kp for this equilibrium

I still need help

Answer 1

EDIT: Additional explanation added.

In one liter of the gas, the total no of moles of gaseous substance is found from the gas law

PV=nRT, where n= no of moles [The pressure/volume/temperature relation depends on the total no of moles, not on what kind, and they can be mixed.]

n = PV/RT; plug in P = 0.155 atm, V = 1 liter, T = 303ºK, R = 8.314 joule/ºK*mole

n = 0.00623 moles in one liter. There are three molecular components in the gas: one CO2 molecule and 2 NH3 molecules, so the no of moles for each component is n/3, or
0.00208. Therefore

[CO2] = 0.00208 and [NH3] = 2 x 0.0028 = 0.00416.

The equilibrium constant is defined as

kp = [CO2]*[NH3]*[NH3] / [N2H6CO2]

[Square brackets indicate the concentrations (moles per liter) of each compound. The NH3 appears twice because the dissociation produces 2 molecules of NH3. The compound 2H6CO2 remains in solid form in equilibrium with the gaseous components; this is analagous to the solubility product of a saturated solution in equilibrium with undissolved solute, for which the concentration of the solute is taken as unity.]

kp = [CO2]*[NH3]^2

kp = 3.59*10^-8