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In a mixture of He, Ne, and Xe gases with a total pressure of 925 atm, if there is 10.5 g of each gas in the mixture, what is the partial pressure of Xe?

Respuesta :

Dejanras
Answer is: partial pressure of Xe is 22,95 atm.
m(He) = 10,5 g.
n(He) = m(He) ÷ M(He).
n(He) = 10,5 g ÷ 4 g/mol
n(He) = 2,625 mol.
m(Ne) = 10,5 g.
n(Ne) = m(Ne) ÷ M(Ne).
n(Ne) = 10,5 g ÷ 20,18 g/mol.
n(Ne) = 0,52 mol.
m(Xe) = 10,5 g.
n(Xe) = 10,5 g ÷ 131,3 g/mol.
n(Xe) = 0,08 mol.
Using Dalton's law:
p(Xe) = (0,08 mol / 0,08 mol + 0,52 mol + 2,625 mol) · 925 atm.
p(Xe) =  22,95 atm.

Answer:

Partial pressure of Xe = 22.9 atm

Explanation:

Given:

Total pressure of gas mixture = 925 atm

Mass of He = Ne = Xe = 10.5 g

To determine:

Partial pressure of Xe

Explanation:

As per Dalton's Law in a mixture of gases the total pressure is the equal to the sum of partial pressures

In this case:

[tex]Ptotal = p(Xe) + p(He) + p(Ne)\\[/tex]-------------(1)

Partial pressure of each gas can be expressed as:

[tex]Partial\ pressure = mole\ fraction * total\ pressure[/tex]

[tex]pXe = \frac{n(Xe)}{n(total)} * Ptotal[/tex]-----(2)

where n(Xe) = moles of Xe

n(Total) = total moles

[tex]nXe = \frac{mass\ Xe}{At\ mass\ Xe} = \frac{10.5g}{131g/mol} =0.080[/tex]

[tex]nHe = \frac{mass\ He}{At\ mass\ He} = \frac{10.5g}{4g/mol} =2.625[/tex]

[tex]nNe = \frac{mass\ Ne}{At\ mass\ Ne} = \frac{10.5g}{20g/mol} =0.525[/tex]

Therefore,

n(Total) = 0.080+2.625+0.525 = 3.23

Substituting for nXe, n(Total) and P(total) in equation (2)

[tex]pXe = \frac{0.080}{3.23} * 925 = 22.9 atm [/tex]