Answer to Question #294684 in General Chemistry for abby

Question #294684

A flask contains a mixture of two gases: NO2 and Ne. If the flask contains 0.768 mol of NO2 and 14.8 g of Ne at a total pressure of 1.17 bar, what is the partial pressure of Ne in the flask (in bar?)

TIP: You don't need the volume of the flask nor the temperature to solve this problem.


1
Expert's answer
2022-02-08T10:00:07-0500

n = m / Mr

n(Ne) = m(Ne) / Mr(Ne) = 14.8 g / 20.2 g/mol = 0.732 mol

n(total) = n(Ne) + n(NO2) = 0.732 mol + 0.768 mol = 1.5 mol

"\\chi"(Ne) = n(Ne) / n(total) = 0.732 mol / 1.5 mol = 0.488

PA = "\\chi"A Ptotal

PNe = "\\chi"Ne Ptotal = 0.488 x 1.17bar = 0.57 bar


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