Question

In a mixture of He, Ne, and Xe gases with a total pressure of 925 atm,...

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?

Homework Answers

Answer #1

we know that

moles = mass / molar mass

so

moles of He = 10.5 g / ( 4 g/mol) = 2.625 mol

moles of Ne = 10.5 g / ( 20 g/mol) = 0.525 mol

moles of Xe = 10.5 g / (131.293 g/mol) = 0.08 mol

now

total moles of gases = 2.625 + 0.525 + 0.08 = 3.23 mol

now

mole fraction of Xe = moles of Xe / total moles of gases

mole fraction of Xe = 0.08 mol / 3.23 mol

mole fraction of Xe = 0.024768 mol

now

partial pressure of Xe = mole fraction of Xe x total pressure

partial pressure of Xe = 0.024768 x 925 atm

partial pressure of Xe = 22.91 atm

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