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