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 5.00 atm . Calculate the partial pressure of each gas in the container.
Express the pressure values numerically in atmospheres, separated by commas. Enter the partial pressure of methane first, then ethane, then propane.
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Expert's answer
2017-01-26T05:07:57-0500
n = m/M
M (CH 4 ) = 16 g/mol M (C 2 H 6 ) = 30 g/mol
n (CH 4 ) = 8/16 = 0.5 mole
n (C 2 H 6 ) = 18/30 = 0.6 mole
PV = nRT
P = 3.6 atm V = 10 L R = 0.0821 L atm/K/mol T = 23 + 273 = 296 K
3.6 · 10 = n · 0.0821 · 296
36 = n · 24.302
n = 36 / 24.302 = 1.481 moles
Total no.of moles = 0.5 + 0.6 + n (C 3 H 8 ) = 1.481 moles
n (C 3 H 8 ) = 1.481 - 1.1 = 0.381 mol
Mole fraction of CH 4 = no.of moles of CH 4 /total no.of moles = 0.5/1.481 = 0.338
Mole fraction of ethane = 0.6/1.481 = 0.405
Mole fraction of propane = 0.381/1.481 = 0.257
Partial pressure of CH 4 = mole fraction of CH 4 · total pressure = 0.338 · 3.6 = 1.216 atm
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