Troubleshooting Mass Flow Rates for CH4 Reaction with Air and Fuel

[Jacob87411]

I have this reaction:

CH4+4(O2+3.76) --> CO2+2H2O+2O2+15.04N2

Through other calculations and given information I have found that the mass flow rate of the air (the O2+3.76N2) is 25kg/s and the mass flow rate of the fuel is .87 kg/s. I am having the hardest time finding, using these mass flow rates, the number of moles of each product I have. Any help is appreciated

 

[mezarashi]

Mass flow rates are given in kg/second, and the similarly you would get moles/second when converted into moles.

How would you normally convert 1kg of O2 into moles of O2?

Hint: 1kg/sec x moles/kg = moles/sec

 

[Jacob87411]

So if I have 25 kg/s of air and air is 21% O2 I have (.21*25kg/s) gives kg/s of O2. Then (.21*25 kg/s)/32 = N_O2? Similarly ((.79*25kg/s)(3.76N2))/28 gives N_N2

How would it work for CO2 for example? Do I find the amount of moles of C, then use the fact that 25% of the moles of oxygen are paired with the C?

 

[mezarashi]

Get the kilograms/sec of each molecule using the mass percentages. Then convert it into moles/sec using that substance's molar mass.

For example the molecule CO2 has a molar mass of (C)12.01+(O)16.00+(O)16.00=44.01 grams per mole.

 

[Jacob87411]

Ok this is what I got (I really appreciate the help by the way):
CO2=44 g/mole
H2O=18 g/mole
O2=32 g/mole
N2=28 g/mole

Total mass of exhaust=565/12
massfractionCO2=.077
massfractionH2O=.064
massfractionO2=.113
massfractioN2=.745

Mass flow of exhaust=mass flow air + mass flow fuel = 25.87 so i take the mass fractions from above and multiply them by that and they give:

CO2=2.00kg/s
H2O=.1.65kg/s
O2=2.92kg/s
N2=19.27 kg/s

And I know for example 1 mole of CO2=44 grams so 2.00 kg/s = 45.45 moles of CO2/s? Is this all correct?

 

[mezarashi]

Just make sure you carefully check each of your mass flow results for each substance (as I haven't checked them). The concept is correct, and yes 2kg/s of CO2 is equivalent to approx 45 moles/s of CO2. =)