Answer to Question #97302 in General Chemistry for Andre W

Question

Three gases (8.00 g of methane, CH4, 18.0 g of ethane, C2H6, and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 ∘C, the total pressure in the container is 4.40 atm . Calculate the partial pressure of each gas in the container.

Answer 1

n1=8/16= 0.5 (CH4)

n2= 18/30= 0.6 (C2H4).

pV=nRT

n=pV/RT = 4.4x10/0.082 x 300= 1.789 moles.(total)

ntotal=n1+n2+n3.

n3=ntotal-n1-n2=1.789-0.5-0.6=

0.689 mol(C3H8).

p1=4.4 x 0.5= 2.2 atm.

p2=4.4 x 0.6=2.64 atm.

p3=4.4 x 0.689= 3.0316atm