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?

Answer 1

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.

Answer 2

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.

Answer 3

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.