NH4HS<-->NH3+H2S

Question

What is the minimum mass of NH4HS that must be added to the 5.00-L flask when charged with the 0.300g of pure H2S, at 25C to achieve equilibrium??

Kp=.120 at 25C
initial pressure of H2S in flask = .0431atm
partial pressures of H2S and NH3
.326, .369
mole fraction of H2S =.531

I used the ideal gas law but still receive the wrong answer. Anyone can help please?

Answer 1

after using the ideal gas law to find moles then mass of NH3 to add to the H2S I got a answer of 3.0334g
This is my last attempt to enter in a answer for the program I am using and I don't want to lose points, obviously. I just want to see if anyone can check to make sure I am doing this right

Answer 2

What did you do with the ideal gas law?
I think you can plug the pressure of NH3 into PV = nRT, solve for n, convert to grams NH4HS. That should do it.

Answer 3

I plugged in the values P is pressure I have for NH3 which is 2.78, 5L for V, n, R the gas constant and T 298K. to get .5684 mol to .03388 g of NH3

Answer 4

If I do it the PV = nRT way I get 3.40 g NH4HS. If I do it by adding grams, I get
0.06655 x 17 = 1.13 g NH3
0.06655 x 34 = 2.26 g H2S
for a total of 3.39.
Looks pretty convincing to me but you should make sure of the s.f. and be sure the rounding is ok.

Answer 5

Thank you for taking the time to answer me on this. I used the gaw law to solve for NH3 for the mass of it to then add it to the mass of H2S to find NH4HS. I think Im still a little confused on how to go about this, but hopefully by your answer I can figure it out.

Answer 6

I know I told you last night to add the grams but it's easier to go with the gas law and convert to g NH4HS.
PV = nRT and use p for NH3; that gives you n for NH3.
moles NH3 = moles NH4HS
g NH4HS = mols NH4HS x molar mass NH4HS.