Question

A mixture containing 2.65 g each of CH4(g), C2H4(g) and C4H10(g) is contained in a 1.50...

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.

Homework Answers

Answer #1

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

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