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?
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.
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.
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?
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.
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.
Nevermind I didn't realize 1380 was stated in the problem, but I still not sure which answer choice to choose.
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.
Ok so the answer has to be in mm Hg since the total pressure was in mm Hg correct?
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.
oh ok because the answer of 0.258 threw me off but I understand why that's not the answer.
Yes, you have correctly calculated the moles of CO2 and CO using the molar mass conversion factor.
For CO2:
56 g CO2 * (1 mol CO2 / 44.01 g CO2) = 1.27 mol CO2
For CO:
3.04 mol CO * (1 mol CO / 28.01 g CO) = 0.108 mol CO
Now, let's check your calculation for the moles of Argon.
For Argon:
9.03 x 10^23 atoms of Ar * (1 mol Ar / 6.022 x 10^23 atoms) = 1.5 mol Ar
So the correct value for the moles of Argon is 1.5 mol, not 2.26 x 10^22 mol as you calculated.
Now, to find the partial pressure of Argon, you can use the ideal gas law equation:
PV = nRT
P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we only need to find the partial pressure of Argon, we can use the equation:
P_Ar = (n_Ar / n_total) * P_total
Substituting the given values:
P_Ar = (1.5 mol / (1.27 mol + 0.108 mol + 1.5 mol)) * 1380 mm Hg
Calculating the denominator:
1.27 mol + 0.108 mol + 1.5 mol = 2.878 mol
Now, substitute into the partial pressure equation:
P_Ar = (1.5 mol / 2.878 mol) * 1380 mm Hg
P_Ar = 719.5 mm Hg
Therefore, the partial pressure of Argon in the mixture of gases is approximately 719.5 mm Hg.