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?
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.
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.
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.
To find the partial pressure of argon in the mixture, you can use Dalton's law of partial pressures. According to the law, the total pressure of a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases.
First, let's calculate the moles of each gas in the mixture:
1. For CO2:
Given: Mass of CO2 = 56 grams
To get moles: Divide the mass by the molar mass of CO2 (44 g/mol)
Moles of CO2 = 56 g / 44 g/mol ≈ 1.27 mol
2. For CO:
Given: Moles of CO = 3.04 mol
3. For Argon (Ar):
Given: Number of atoms of Ar = 9.03 × 10^23 atoms
To get moles: Divide the number of atoms by Avogadro's number (6.022 × 10^23 mol⁻¹)
Moles of Ar = 9.03 × 10^23 atoms / 6.022 × 10^23 mol⁻¹ ≈ 1.50 mol
Next, sum up the moles of all the gases in the mixture:
1.27 mol (CO2) + 3.04 mol (CO) + 1.50 mol (Ar) ≈ 5.81 mol
Now, to find the partial pressure of argon, we need to know the total pressure of the mixture. Given that the total pressure is 1380 mm Hg, we can use the following formula for partial pressure:
Partial pressure = (Moles of gas / Total moles of all gases) × Total pressure
Partial pressure of Ar = (1.50 mol / 5.81 mol) × 1380 mm Hg
Partial pressure of Ar ≈ 355.90 mm Hg
Therefore, the partial pressure of argon in the mixture is approximately 355.90 mm Hg.