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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?

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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.

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