Amixture of methane gas, ch4(g) , and pentane gas, c5h12(g) , has a pressure of 0.5239 atm when placed in a sealed container. the complete combustion of the mixture to carbon dioxide gas, co2(g) , and water vapor, h2o(g) , was achieved by adding exactly enough oxygen gas, o2(g) , to the container. the pressure of the product mixture in the sealed container is 2.429 atm . calculate the mole fraction of methane in the initial mixture, assuming the temperature and volume remain constant.

Let the partial pressure of methane, CH₄ be u

Let the partial pressure of pentane, C₅H₁₂ be v

The total pressure of the mixture is 0.5239 atm i.e

Partial pressure of methane + partial pressure of pentane = 0.5239 atm

Combustion reaction of the mixture.

CH₄ + 2O₂ —> CO₂ + 2H₂O

From the balanced equation above,

u atm of CH₄ produced u atm of CO₂ and 2u atm of H₂O.

Thus, the partial pressure of CH₄ produces = u + 2u = 3u atm of the products.

C₅H₁₂ + 8O₂ —> 5CO₂ + 6H₂O

From the balanced equation above,

v atm of C₅H₁₂ produced v atm of CO₂ and 6v atm of H₂O.

Thus, the partial pressure of C₅H₁₂ produces = 5v + 6v = 11v atm of the products.

CH₄ produced 3u atm

C₅H₁₂ produced 11v atm

But the total pressure of the products is 2.429 atm.

Thus,

Next, we shall determine the partial pressure of methane, CH₄ i.e u.

u + v = 0.5239 1

3u + 11v = 2.429 2

From equation 1,

v = 0.5239 – u

Substitute v into equation 2

3u + 11v = 2.429

3u + 11(0.5239 – u) = 2.429

3u + 5.7629 – 11u = 2.429

Collect like terms

3u – 11u = 2.429 – 5.7629

–8u = –3.3339

Divide both side by –8

u = –3.3339 / –8

Therefore, the partial pressure of methane, CH₄ in the initial mixture is 0.4167 atm.

Finally, we shall determine the mole fraction of methane CH₄ in the initial mixture.

Partial pressure of methane, CH₄ = 0.4167 atm.

Total pressure = 0.5239 atm

Mole fraction = partial pressure / Total pressure

Mole fraction of methane CH₄ = 0.4167 / 0.5239

Therefore, the mole fraction of methane CH₄ in the initial mixture is 0.795

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