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
