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
