Question

(a) A mixture containing 2.15 g each of CH₄(g), C₂H₄(g) and C₄H₁₀(g) is contained in a 1.50 L flask at a temperature of 25°C.

Calculate the partial pressure of each of the gases in the mixture.

PCH4	= ___ atm
PC2H4	= ___ atm
PC4H10	= ___ atm

Calculate the total pressure of the mixture.
___ atm

(b) A mixture of gases contains 1.06 g of N₂, 2.22 g of H₂, and 1.22 g of NH₃. If the total pressure of the mixture is 4.12 atm, what is the partial pressure of each component?

PN2	= ___ atm
PH2	= ___ atm
PNH3	= ___ atm

Answer 1

a) Given that a mixture containing 2.15 g each of CH4(g), C2H4(g) and C4H10(g) is contained in a 1.50 L flask at a temperature of 25°C.

Ideal gas equation is PV = nRT

where n = total moles = sum of moles of individual components

P = total pressure of the mixure

So first we have to find out the total moles of the mixture to get total pressure of the mixure.

Moles of CH₄ = mass of CH4/ molar mass of CH4 = 2.15 g/ 16 gmol-1 = 0.134 moles

    Moles of C₂H₄ = mass of C2H4/ molar mass of C2H4 = 2.15 g/ 28 gmol-1 = 0.077 moles

   Moles of C₄H₁₀ = mass of C4H10/ molar mass of C4H10 = 2.15 g/ 58 gmol-1 = 0.037 moles

Total moles =  Moles of CH₄ + Moles of C₂H₄ +Moles of C₄H₁₀

   =0.134 moles +0.077 moles + 0.037 moles

= 0.248 moles

Total moles n = 0.248 moles

Given that volume of flask V = 1.50 L

temperature T = 25^oC = 25 +273 K= 298 K

PV = nRT

P = nRT/ V

= (0.248 mol) (0.0821 L.atm/K/mol) (298 K) / (1.50 L)

= 4.045 atm

P = 4.045 atm

Therefore, total pressure of the mixture = 4.045 atm

Partial pressure of CH4

Partial pressure of CH4 = (moles of CH4/ total moles of the mixture) x total pressure of the mixture

= (0.134/0.248) x 4.045 atm

= 2.185 atm

Hence, Partial pressure of CH4 = 2.185 atm

Partial pressure of C2H4

Partial pressure of C2H4 = (moles of C2H4/ total moles of the mixture) x total pressure of the mixture

= (0.077/0.248) x 4.045 atm

= 1.256 atm

Hence, Partial pressure of C2H4 = 1.256 atm

Partial pressure of C4H10

Partial pressure of C4H10 = (moles of C4H10/ total moles of the mixture) x total pressure of the mixture

= (0.037/0.248) x 4.045 atm

= 0.603 atm

Hence, Partial pressure of C4H10 = 0.603 atm

b) Given that a mixture of gases contains 1.06 g of N2, 2.22 g of H2, and 1.22 g of NH3.

Also given that total pressure of the mixture is 4.12 atm.

Partial pressure of gas A = (moles of A / total moles of the mixture) x total pressure of the mixture

So we have to find out, individual moles of each gas present in the mixture.

Moles of N2= mass of N2/ molar mass of N2 = 1.06 g/ 28 gmol-1 = 0.034 moles

    Moles of H2= mass of H2/ molar mass of H2 = 2.22 g/ 2 gmol-1 = 1.11 moles

   Moles of NH3= mass of NH3/ molar mass of NH3 = 1.22 g/ 17 gmol-1 = 0.072 moles

Total moles =  Moles of N2+ Moles of H2+Moles of NH3

   =0.034 moles +1.11 moles + 0.072 moles

= 1.216 moles

Total moles = 1.216 moles

Partial pressure of N2

Partial pressure of N2 = (moles of N2/ total moles of the mixture) x total pressure of the mixture

= (0.034/1.216) x 4.12 atm

= 0.115 atm

Hence, Partial pressure of N2 = 0.115 atm

Partial pressure of H2

Partial pressure of H2 = (moles of H2/ total moles of the mixture) x total pressure of the mixture

= (1.11/1.216) x 4.12 atm

= 3.76 atm

Hence, Partial pressure of H2 = 3.76 atm

Partial pressure of NH3

Partial pressure of NH3 = (moles of NH3/ total moles of the mixture) x total pressure of the mixture

= (0.072/1.216) x 4.12 atm

= 0.244 atm

Hence, Partial pressure of NH3 = 0.244 atm