A mixture of gases contains twice as many moles of As as Kr. If 0.300 mol of Xe is added to this mixture, the pressure increases from 1.26 atm to 1.47 atm. how many moles of Ar are in the mixture?
If there are x moles of Kr, then there are 2x moles of Ar. Since the number of moles has increased by a factor of 1.47/1.26,
2x+x+.3 = 3x(1.47/1.26)
x = 0.6
So, there were originally 1.2 moles of Ar, making the final amount 1.5 moles.
Steve has worked this as a math problem (because he is a math person). I assume the title of chemistry for the question you may want to work it as a chemistry problem. Do it this way. BTW, I agree with Steve's answer EXCEPT for the 1.5 mols. If mols Ar are 1.2 in the initial mixture it must still be 1.2 in the final.
pAr + pKr =1.26 atm
pAr + pKr + pXe =1.47 atm
Therefore, pXe = 1.47-1.26 = 0.21 atm.
PV = nRT and since V, R, and T are constant in this problem, we can rewrite PV = nRT as P = nk where all of the constant stuff is in that k. Then 0.21 = 0.3k and k = 0.7
Again, P = nk and 1.47 = n(0.7) and n = 1.47/0.7 = 2.1 and that's total mols of Ar, Kr, Xe. Since we know Xe is 0.3, then Ar + Kr must be 2.1-0.3 = 1.8
Then we know Ar + 1/2 Ar = 1.8 so
Ar must be 1.8/1.5 = 1.2 in the mixture (both the original mixture as well as the final mixture). Total mols in the original mixture is 1.8 and total in the final is 2.1.
I agree with almost all of your work. I also would agree that working it as a math problem is a lot easier than as a chemistry problem. I did it as a chem problem because it was labeled chem. While the mols Ar = 1.2 it is 1.2 both in the initial mixture as well as the final mixture. That extra 0.3 mol added to make the second mixture was Xe and not Ar or Kr.
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = gas constant
T = temperature
We're given that the pressure increases from 1.26 atm to 1.47 atm when 0.300 mol of Xe is added. To find the number of moles of Ar in the mixture, we can use the properties of an ideal gas to calculate the change in moles of Xe.
First, let's find the initial number of moles of Xe in the mixture. Since the volume and temperature are constant, we can use the equation:
P1V1 = n1RT
Given:
P1 = 1.26 atm
V1 = volume (assuming constant)
R = gas constant
Next, we'll find the final number of moles of Xe in the mixture using the same equation:
P2V2 = n2RT
Given:
P2 = 1.47 atm
V2 = volume (assuming constant)
R = gas constant
Subtracting the initial number of moles from the final number of moles will give us the change in moles of Xe.
Finally, since the number of moles of As is twice the number of moles of Kr, we can find the number of moles of Kr and then calculate the number of moles of Ar.