View?offset nextBassignmentProblemiD 94246909 2 of 12 Three gases (8 00 g of mat

Question

View?offset nextBassignmentProblemiD 94246909 2 of 12 Three gases (8 00 g of mathane, CHL, 18 0 g of othan, Cll and an unknown amount of propane Cill wore ashed to the sae 10 0 L. container Al 23 0 C the total pressure in the container is ts 30 atm Calculale the partial pressure of each gus in Express the pressure values numerically in atmospheres then ethane, then propane. thhe contane separated by commas Enter the partial pressure of methane first, View Available Hint(s) atm Submit Part B A gaseous mixture of 02 and N2 contans 32 8 % ntrogen by mass whats the patal pressure of oxypern hmt o fentdal pressure is 645 mmHg? Express you answer numerically in millimeters of mereury View Available Hints) mmHg

Explanation / Answer

A) partial pressure = mole fraction x total pressure...

moles CH4 = 8.00 g x (1 mole / 16.0 g) = 0.500 moles
moles C2H6 = 18.0 g x ( 1mole / 30.0 g) = 0.600 moles

next.. find total moles...
PV = nRT
n = PV/RT = (5.30 atm) x (10.0 L) / [ (0.0821 Latm/moleK) x (296K) ] = 2.18 moles

moles of propane = 2.18 - 0.500 - 0.600 = 1.08 moles

partial pressure methane = (0.500 / 2.18) x 5.3 atm = 1.22 atm
partial pressure ethane = (0.600 / 2.18) x 5.3 atm = 1.46 atm
partial pressure propane = 5.3 - 1.22 - 1.46 = 2.62 atm.

B)Basis: We take 100 g of mixture.Then,

We have 32.8 / 28 = 1.17 mole N2 and 67.2 / 32 = 2.1 moles O2
Total moles = 1.17 + 2.1 = 3.27
nA / n T = pA / pT
1.17 / 3.27 = pN2 / 645
p N2 = 230.78 mmHg

2.1 /3.27 = pO2 / 645
pO2 = 414.22 mm Hg

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