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?

Question

Chemistry

Emery Conrad

19 January, 08:39

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?

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Answers (1)

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Juliette Warner · Answer 1

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.