A mixture containing 2.65 g each of CH4(g), C2H4(g) and C4H10(g) is contained in a 1.50 L flask at a temperature of 25°C. (a) Calculate the partial pressure of each of the gases in the mixture. PCH4 = atm PC2H4 = atm PC4H10 = atm (b) Calculate the total pressure of the mixture.
mass of each gas = 2.65g
moles of Ch4 = 2.65g/molar mass
= 2.65/16
= 0.1656
moles of C2H4 = 2.65/28
=0.09464
moles of C4H10 = 2.65/58
= 0.04568
Partial pressure of methane = nRT/V
= 0.1656mol x 0.0821 L.atm/mol.K x 298K/ 1.50L
= 2.701 atm
Partial pressure of C2H4 = nRT/V
= 0.09464mol x 0.0821 L.atm/mol.K x 298K/ 1.50L
= 1.543atm
Partial pressure of C4H10 = nRT/V
= 0.04568mol x 0.0821 L.atm/mol.K x 298K/ 1.50L
= 0.7450atm
Total pressure = 0.7450 + 1.543 +2.701
= 4.989 atm
Get Answers For Free
Most questions answered within 1 hours.