Question

Propane fuel (C \(_{3} \mathrm{H}_{8}\) ) is burned with stoichiometric amount of air in a water heater. The products of combustion are at 1 atm pressure and \(120^{\circ} \mathrm{F}\). What fraction of the water vapor in the products is vapor?

Short answer

Answer: The fraction of water vapor formed in the products is approximately 0.5714 or 57.14%.

Step-by-step solution

Write the balanced chemical equation for the combustion of propane

The combustion of propane (C3H8) with air (O2) produces carbon dioxide (CO2) and water vapor (H2O). The balanced chemical equation for this reaction is: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

Calculate the moles of reactants and products in the stoichiometric reaction

According to the balanced chemical equation, for every mole of propane (C3H8), 5 moles of oxygen (O2) react stoichiometrically to produce 3 moles of carbon dioxide (CO2) and 4 moles of water vapor (H2O). Assume one mole of propane is burned in the water heater: 1 mol C3H8 + 5 mol O2 → 3 mol CO2 + 4 mol H2O

Calculate the total moles of products

The total moles of products formed in the reaction are equal to the sum of moles of carbon dioxide and water vapor: Total moles of products = moles of CO2 + moles of H2O Total moles of products = 3 mol + 4 mol = 7 mol

Calculate the mole fraction of water vapor

The mole fraction of water vapor (H2O) in the products is the ratio of moles of water vapor to the total moles of products: Mole fraction of H2O = moles of H2O / total moles of products Mole fraction of H2O = 4 mol / 7 mol Mole fraction of H2O = 4/7 ≈ 0.5714 So, the fraction of water vapor in the products is approximately 0.5714 or 57.14%.