A mixture of three gases has a total pressure of 1380 mm Hg at 25 degrees celsius. If the mixture contained 56 grams CO2, 3.04 mol CO and 9.03 X 10^23 atoms of Argon, the partial pressure of argon would be about?

Question

First I did 9.03e23 * 760/1 = 2.26e22 and then I was going to do that answer * 0.0821 ans the temp which is 298 and then divide by the volume but then I realized they didn't give volume and now im stuck.

Answer 1

I would convert 56 g CO2 to moles, convert 9.03E23 atoms to moles, then find mole fraction Ar which will be
n argon/total n = XAr.
Then pAr = XAr x total P.
Don't forget that you hav 3.04 mol CO.

Answer 2

ok so first I did 56 CO2 X 1/44.01 = 1.27

then 3.04 CO X 1/28.01 = 0.108

then 9.03e23 Ar X 1/39.95 = 2.26e22 but im not sure I did this right.

Is this correct so far before I continue? Thank you for your help.