What is the minimum mass of nh4hs that must be added to the 5.00-l flask when charged with the 0.250 g of pure h2s(g), at 25 ∘c to achieve equilibrium?

We first can get the No of moles of H2S = 0.25 x 1 mol(H2S) / 34 g (H2S)
                                                                   = 0.0074 Moles
and according to the ideal gas low:
we can get p for H2S
PV = nRT
when we have V = 5 L & n = 0.0074  & R (constant)= 0.082 & T= 25 + 273 = 298 K
By substitution:

P* (5L) = (0.0074)*(0.082)*(298)
∴ P = 0.036 atm
By assuming Kp (should be given, you just missed it) = 0.12 at 25 C°
By substitution: to get P for NH3

0.12 = X ( P + X)
0.12 = X ( 0.036 + X)
∴X^2 + 0.036 X -0.12 = 0
by solving this equation we get 
X= 0.365 atm
So to get the no of moles of NH3:
PV = nRT
0.365 * 5 = n ( 0.082*298)
∴ n = 0.075 moles 
and to get the mass on (g) =no.of moles * molar mass
                                                      0.075 * 51 = 3.825 g NH4Hs

Tolerance to high levels of salinity