Question

A sample of a gas mixture contains the following quantities of three gases. compound mass CO 3.13 g CO2 3.87 g SF6 2.01 g The sample has: volume = 2.50 L temperature = 16.6 °C What is the partial pressure for each gas, in mmHg? What is the total pressure in the flask?

Answer 1

1) CO:

Molar mass of CO,

MM = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

mass(CO)= 3.13 g

use:

number of mol of CO,

n = mass of CO/molar mass of CO

=(3.13 g)/(28.01 g/mol)

= 0.1117 mol

Given:

V = 2.5 L

n = 0.1117 mol

T = 16.6 oC

= (16.6+273) K

= 289.6 K

use:

P * V = n*R*T

P * 2.5 L = 0.1117 mol* 0.08206 atm.L/mol.K * 289.6 K

P = 1.0618 atm

P = 1.0618*760 mmHg

= 807 mmHg

Answer: 807 mmHg

2) CO2

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass(CO2)= 3.87 g

use:

number of mol of CO2,

n = mass of CO2/molar mass of CO2

=(3.87 g)/(44.01 g/mol)

= 8.793*10^-2 mol

Given:

V = 2.5 L

n = 0.0879 mol

T = 16.6 oC

= (16.6+273) K

= 289.6 K

use:

P * V = n*R*T

P * 2.5 L = 0.0879 mol* 0.08206 atm.L/mol.K * 289.6 K

P = 0.8356 atm

P = 0.8356*760 mmHg

= 635 mmHg

Answer: 635 mmHg

3) SF6

Molar mass of SF6,

MM = 1*MM(S) + 6*MM(F)

= 1*32.07 + 6*19.0

= 146.07 g/mol

mass(SF6)= 2.01 g

use:

number of mol of SF6,

n = mass of SF6/molar mass of SF6

=(2.01 g)/(1.461*10^2 g/mol)

= 1.376*10^-2 mol

Given:

V = 2.5 L

n = 0.0138 mol

T = 16.6 oC

= (16.6+273) K

= 289.6 K

use:

P * V = n*R*T

P * 2.5 L = 0.0138 mol* 0.08206 atm.L/mol.K * 289.6 K

P = 0.1312 atm

= 0.1312*760 mmHg

= 99.7 mmHg

Answer: 99.7 mmHg

4)

P Total = sum of all pressures

= 807 mmHg + 635 mmHg + 99.7 mmHg

= 1.54*10^3 mmHg

Answer: 1.54*10^3 mmHg