A certain mixture of helium, He, and krypton, Kr, has a mass of 10.0 g and occupies a volume of 20.0 L at 25.0EC and 1.20 atm. What is the mass of He in the mixture?

Answer 1

Let X = mass He
and Y = mass Kr.
Two equations and two unknowns.
eqn 1 is X + Y = 10
Then use PV = nRT and solve for n = total mols.
equation 2 is
(X/4) + (Y/83.8) = total mols

Solve for X = grams He.

Answer 2

I used the ideal gas law, and I got .980456 moles. then I used the equation to substitute Helium=10.0g -Krypton in the equation (x/4)+(y/83.8)=. 980456 mols for X. With a little bit of algebra, I got 6.38079g of Krypton and 3.61921g of Helium. Thanks you for the help. I wuld of never got the second equation by myself. I think I did this correctly, if not I would really appreciate someone responding.

Answer 3

I went through the calculations quickly and obtained 3.622 which rounds to 3.62 for He. The small difference in our numbers probably comes from my obtaining 0.981 for n. I'm positive you worked it correctly.