0.0100 mol of NH4Cl and 0.0100 mol of NH3 are placed in a closed 2.00 L container and heated to 603K. At this

0.0100 mol of NH4Cl and 0.0100 mol of NH3 are placed in a closed 2.00 L container and heated to 603K. At this temperature , all the NH4Cl vaporizes.

NH4Cl (s) ↔ NH3 (g) + HCl (g)
When the reaction has come to equilibrium, 5.8 x 10-3 mol of HCl is present. Calculate
a) Equilibrium constant Kc

Ok so I get that the Kc = [NH3][HCl] because the NH4Cl is a solid and those are not part of the Kc. I also know that I need to set up an ice table, but when I set it up I get hung up on the actual ice table itself. The ice table only consists of the products, so would the change be negative for NH3 (since that is what was placed into the system) and positive for HCl since there was none in the system?if so does my equation turn into:

Kc = [0.0100-x][+x] ?
And if it is how in the hell would I solve this?

0.0100 mol of NH4Cl and 0.0100 mol of NH3 are placed in a closed 2.00 L
These need to be concentrations so divide by 2L ===> 0.0050 M

I cannot draw a table here ... so I'll fake it
5.8 x 10-3 mol of HCl is present at equilib.

NH4Cl (s) ↔ NH3 (g) + HCl (g) ---- notice all coefficients are 1 so all changes are x
I= 0.0050 ... 0.0050 .... 0
C= -x ..... x ..... x
E = 0.0050 -x ... 0.0050 +x... x ===> x is NOT an unknown,
but is a concentration 5.8 x 10-3 mol / 2L ===> x = 2.9 x 10-3 M
So Kc = [0.0050 + 2.9 x 10-3][2.9 x 10-3] = [7.9 x 10^-3][2.9 x 10-3]
Kc = 2.291 x 10^-5