NH4HS<-->NH3+H2S
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?
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.
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
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.
To solve this problem, we need to use the ideal gas law and the expression for the equilibrium constant (Kp).
First, let's calculate the number of moles of H2S originally present in the flask:
n(H2S) = mass(H2S) / molar mass(H2S)
Given:
mass(H2S) = 0.300 g
molar mass(H2S) = ?
We need to find the molar mass of H2S. The molar mass of hydrogen (H) is approximately 1 g/mol, and the molar mass of sulfur (S) is approximately 32 g/mol. Therefore, the molar mass of H2S is:
molar mass(H2S) = (2 * molar mass(H)) + molar mass(S) = (2 * 1 g/mol) + 32 g/mol = 34 g/mol
Now we can calculate the number of moles of H2S:
n(H2S) = 0.300 g / 34 g/mol
Next, we need to calculate the volume of the flask in moles using the ideal gas law:
PV = nRT
Given:
P = pressure = 0.0431 atm
V = volume = 5.00 L
T = temperature = 25°C = 298 K (convert Celsius to Kelvin)
R = ideal gas constant = 0.0821 L·atm/(mol·K)
Rearranging the ideal gas law equation, we can solve for the number of moles (n):
n = PV / RT
Now we can calculate the number of moles of H2S initially present in the flask:
n(H2S) = (0.0431 atm) * (5.00 L) / (0.0821 L·atm/(mol·K) * 298 K)
Now, to achieve equilibrium, the reaction NH4HS ⇌ NH3 + H2S needs to reach a certain partial pressure ratio determined by the equilibrium constant (Kp). Given that Kp = 0.120, we can use the equation:
Kp = (P(NH3) * P(H2S)) / (P(NH4HS))
We are given the partial pressures of NH3 and H2S, so the equation becomes:
0.120 = (0.369 * P(H2S)) / P(NH4HS)
Now, we can solve for the unknown partial pressure of NH4HS:
P(NH4HS) = (0.369 * P(H2S)) / 0.120
And since we know the equation: P = nRT / V, we can find the number of moles of NH4HS as follows:
n(NH4HS) = (P(NH4HS) * V) / RT
Finally, we can convert moles to grams to find the minimum mass of NH4HS that must be added to the flask.
Note: Make sure to double-check the calculations and units to avoid any errors in the final answer.