posted by student on .
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
mole fraction of H2S =.531
I used the ideal gas law but still receive the wrong answer. Anyone can help please?
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
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.
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
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.
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.
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.