Propane fuel (C3H8) is burned with 30 percent excess air. Determine the mole fractions of each of the products. Also, calculate the mass of water in the products per unit mass of the fuel and the air-fuel ratio.

Answer:mole fraction of carbon dioxide = 0.429;

mole fraction of water = 0.571

Mass of water in the products = 1.635g

Explanation:

The process of burning an hydrocarbon (example:propane) in air is called combustion. This reactions yields carbon dioxide, water and heat. With this information, we can generate a balanced equation of reaction as below:

C₃H₈ + 5O₂ ↔ 3CO₂ + 4H₂O

Note that there is 30% excess air. This means that there is enough oxgen to burn all the propane available. Thus it is a complete combustion process.

Mole fraction = number of moles / total number of moles of the solution

Number of moles of C0₂ = 3

Number of moles of H₂O = 4

Mole fraction of CO₂ = 3/3+4 = 3/7 = 0.429

Mole fraction of H₂O = 4/3+4=4/7 = 0.571

To calculate the mass of water in the products

Molar mass of C₃H₈ = (3*12.011) + (8*15.999) = 44.067g

1mole of C₃H₈ = 44.067g

using a unit mass of the fuel (propane) = 1g,

Therefore 1g of C₃H₈ = 1/44.067 = 0.02269moles

Looking at the equation of reaction, 1mole of C₃H₈ yields 4moles of water,

Therefore 0.02269moles of C₃H₈ will yield:

0.02269*4 = 1.635g of H₂O