- #1
radwa
- 5
- 0
Will you please help me...
It is required to determine the molal analysis of the products of combustion when C8 H18 is burned with 200% theoretical air, and determine the dew point (saturation temperature corresponding to the partial pressure of water present in this equation) of the products if the pressure is 0.1 MPa
This is the combustion equation the book used to solve this problem with
C8 H18 + 12.5(2)O2 +12.5(2)(3.76)N2 =8CO2 +9H2 O+12.5O2 +94N2
My question is when balancing this equation we might have chosen to write it as
C8 H18 + 6.75(2)O2 +6.75(2)(3.76)N2 =8CO2 +9H2 O+O2 +50.76N2
But if we changed the equation like this, in getting the partial pressure of water, instead of calculating it by obtaining the mole percent of the water from the first equation which is 7.29% and multiplying it by 0.1 MPa to have apartial pressure of 7.29 KPa we will get a very different result using the second equation .How do we know that we have to write the first and not the second equation?
It is required to determine the molal analysis of the products of combustion when C8 H18 is burned with 200% theoretical air, and determine the dew point (saturation temperature corresponding to the partial pressure of water present in this equation) of the products if the pressure is 0.1 MPa
This is the combustion equation the book used to solve this problem with
C8 H18 + 12.5(2)O2 +12.5(2)(3.76)N2 =8CO2 +9H2 O+12.5O2 +94N2
My question is when balancing this equation we might have chosen to write it as
C8 H18 + 6.75(2)O2 +6.75(2)(3.76)N2 =8CO2 +9H2 O+O2 +50.76N2
But if we changed the equation like this, in getting the partial pressure of water, instead of calculating it by obtaining the mole percent of the water from the first equation which is 7.29% and multiplying it by 0.1 MPa to have apartial pressure of 7.29 KPa we will get a very different result using the second equation .How do we know that we have to write the first and not the second equation?