A mixture of 5.0 g of Ar and 10.0 g Ne have a total pressure of 2.6 bar. What is the partial pressure of Ar?

Question

m(Ar) = 5.0g , M(Ar) = 39.95g/mol
m(Ne) =10.0g , M(Ne) = 20.18g/mol
n(Ar) = 5.0g/39.95g/mol = 0.125 mol
n(Ne) = 10.0g/20.18g/mol = 0.496 mol
N(total) = 0.125 mol + 0.496 mol = 0.621 mol
X(Ar) = n(Ar)/n(total) = 0.201 %
P(Ar) = X(Ar) * P(total) = (0.201)(2.6 bar) = 0.52 bar

Is my answer right for the given question, also is there a faster way of doing a question like this?

A gas mixture consists of H2, O2, and Ar, where the mole fraction of H2 is 0.35 and the mole fraction of O2 is 0.45. If the mixture is at STP in a 3.0 L container, how many molecules of Ar are present?

For this given question I'm confused onto how to do it. I understand that the total mole fraction must equal to 1, so Ar has the mole fraction of 0.20, but after that I'm confused. Also pressure is 1 bar for STP.

Answer 1

For the 1st question I don't think there is a shorter way of doing it. I didn't check your math but the process looks good EXCEPT for the following:
X(Ar) = n(Ar)/n(total) = 0.201 %
X(Ar) = 0.201. It isn't a percent until multiplied by 100.
I didn't have time to look at the second problem.

Answer 2

So PV = nRT. You know P, V = 3.0 L, and you know R and T. Solve for n = total moles. Multiply by mole fraction Ar (you're 0.20) to find moles Ar, then convert to molecules knowing that 1 mol contains 6.02E23 molecules.