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

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.

Answer 1

mol CO2 is ok.
3.04 mol CO needs no conversion. It is mol in the problem.
9.03E23 atoms Ar x (1 mol Ar atoms/6.022E23 atoms) = mol Ar.

Answer 2

ok so for Ar i got 1.49. Then I added 1.27+1.49+3.04 = 5.8 and then I did 1.49 / 5.8 = 0.256.

Now do I have to multiply 0.256 by the total?

Answer 3

Recalculate n for Ar; it should be 1.50 rounded. I think you threw away the last part. That makes total n = 5.81 and
XAr = 1.50/5.81 = 0.258
Then 0.258*1380 = partial pressure Ar if you want it in mm Hg.

Answer 4

Ok and how did you get 1380? When I did 0.258*1380 I got 356.04.

The answer choices that they gave me are 0.258atm, 300mm Hg, 360 mm Hg, 5300mm Hg, and 8000 mm Hg.

Answer 5

Nevermind I didn't realize 1380 was stated in the problem, but I still not sure which answer choice to choose.

Answer 6

1380 is the total pressure.
PAr = XAr*Ptotal
PAr = 0.258 x 1380 = 356.04 mm Hg which I would round to 356 mm Hg. (356/760= 0.468 atm)
The closest answer seems to be 360 mm Hg.

Answer 7

Ok so the answer has to be in mm Hg since the total pressure was in mm Hg correct?

Answer 8

No, that isn't true. The total prfessure in the problem was 1380 mm Hg but the answer could be in any units. The answer choices you have are all in mm Hg except for the one in atm. That is why I took the answer of 356 mm Hg and divided by 760 to convert 356 mm Hg to atm. That is the 0.468 atm but since 0.468 atm is not a choice, I looked elsewhere for the answer.

Answer 9

oh ok because the answer of 0.258 threw me off but I understand why that's not the answer.