Answer:
a) 2.6 mole of O2 per mole of methane
b) 9.78 mole of N2 per mole of methane
c) Mole fraction of H2O = 0.33
d) 6.12 mole of H2O pero mole of methane
Explanation:
The first thing you have to do is the balanced reaction of the methane combustion:
CH4 +2 O2 -->2 H2O + CO2
You calculate the moles of O2 needed to react with 1 mole of methane and add the excess factor
2 x 1.3 = 2.6 moles of O2
For the N2 moles you calculate it with the air composition (21 % O2, 79% N2)
2.6 moles of O2 * 0.79/0.21 = 9.78 moles of N2
With the total stream of air = 12.38 moles you add the humidity factor
12.38 * 1.5 = 18.57 moles So 6.12 are moles of H20 entering per mole of CH4.
To calculate the mole fraction yo divide the moles of water among the moles of the stream:
6.12/18.57 --> 0.33
