Question

f 2.0 moles of octane vapor, C8H18, is reacted with 15 moles of oxygen to produce carbon dioxide and water, what are the partial pressures of all gases in the mixture at a total pressure of 1 atm?

Answer 1

Write down the balanced chemical equation as

2 C₈H₁₈ (g) +   25 O₂ (g) --------> 16 CO₂ (g) + 18 H₂O (g)

Find out the limiting reactant by finding out which reactant produces a lower amount of the product, say CO₂.

C₈H₁₈: (2 mole C₈H₁₈)*(16 mole CO₂/2 mole C₈H₁₈) = 16 mole CO₂

O₂: (15 mole O₂)*(16 mole CO₂/25 mole O₂) = 9.6 mole CO₂.

Therefore O₂ is the limiting reactant and we can find out the amount of CO₂ produced as 9.6 mole. The amount of H₂O produced = (15 mole O₂)*(18 mole H₂O/25 mole O₂) = 10.8 mole H₂O.

The system contains 2 moles of C₈H₁₈, 15 moles of O₂, 9.6 moles of CO₂ and 10.8 moles of H₂O. Total number of moles of gaseous reactants and products = (2 + 15 + 9.6 + 10.8) moles = 37.4 moles.

Mole fraction C₈H₁₈ = moles C₈H₁₈/total number of moles = 2/37.4 = 0.05347

Mole fraction O₂ = moles O₂/total number of moles = 15/37.4 = 0.40107

Mole fraction CO₂ = moles CO₂/total number of moles = 9.6/37.4 = 0.25668

Moles fraction H₂O = moles H₂O/total number of moles = 10.8/37.4 = 0.28877

Partial pressure of C₈H₁₈ = (mole fraction of C₈H₁₈)*(total pressure) = (0.05347)*(1 atm) = 0.05347 ≈ 0.0535 atm (ans).

Partial pressure of O₂ = (0.40107)*(1 atm) = 0.40107 atm ≈ 0.4011 atm (ans).

Partial pressure of CO₂ = (0.25668)*(1 atm) = 0.25668 atm ≈ 0.2567 atm (ans).

Partial pressure of H₂O = (0.28877)*(1 atm) = 0.28877 atm ≈ 0.2888 atm (ans).