Answer to Question #64807 in General Chemistry for Caroline Boyette

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 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.

Answer 1

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

Partial pressure of ethane = 0.405 · 3.6 = 1.458 atm

Partial pressure of propane = 0.257 · 3.6 = 0.925 atm