calculating partial pressure in gas mixture

Question

A 5.00L

 tank at 4.43°C

 is filled with 5.42g

 of boron trifluoride gas and 8.12g

 of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions.

Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Round each of your answers to 

3

 significant digits.

boron trifluoride	mole fraction:
partial pressure:	atm
sulfur hexafluoride	mole fraction:
partial pressure:	atm
Total pressure in tank:	atm

Answer 1

Boron trifluoride = BF₃

molar mass = 67.82 g/mol

moles present = 5.42 g x 1 mol / 67.82 g = 0.0799 moles BF₃

Sulfur hexafluoride = SF₆

molar mass = 146.1 g/mol

moles present = 8.12 mol x 1 mol / 146.1 g = 0.0556 moles SF₆

Total moles of gas present = 0.0799 mol + 0.0556 mol = 0.1355 moles of gas

Mole fraction BF₃ = 0.0799 / 0.1355 = 0.590

Mole fraction SF₆ = 0.0556 / 0.1355 = 0.410

To find total pressure, we can use the ideal gas law PV = nRT

P = pressure = ?

V = volume = 5.00 L

n = moles = 0.1355 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 4.43ºC + 273.15 = 277.58K

Solving for P we have...

P = nRT/V = (0.1355)(0.0821)(277.58) / 5.00

P = 0.618 atm = total pressure

Partial pressure BF₃ = mole fraction x total pressure = 0.590 x 0.618 atm = 0.365 atm

Partial pressure SF₆ = mole fraction x total pressure = 0.410 x 0.618 atm = 0.253 atm