Material Balance Problem Involving Ideal Gases- Any ?

[ChEtobe14]

Homework Statement

A furnace is fired with 1000 feet cubed per hour at 60 degrees F and 1 atm of a natural gas containing the following volumetric analysis: CH4: 80%, C2H6: 16%, O2: 2%, CO2: 1%, and N2: 1%. The exit flue gas temperature is 800 degrees F and the pressure is 760 mm Hg absolute; 15% excess air is used and combustion is complete. Calculate the volume of the products and the total volume of the flue gas produced.

Homework Equations

pV=nRT
In-Out= (extent of reaction)(stoichiometric coefficient)

The Attempt at a Solution

See attachment for my flow diagram:

Degrees of freedom:

number of unkowns: 5
number of knowns: see defined equations

By stoichiometry:

(1) CH4 + 2O2 ----> CO2 + 2H2O
(2) C2H6 + 7/2O2---->2CO2 + 3H2O

(800 mol CH4)*(2 mol O2/ 1 mol CH4) = 1600 mol O2
(160 mol C2H4)*(7/2 mol O2/1 mol C2H4) = 560 mol O2
Total O2 needed: 2160 mol O2

Excess Air was 15% thus 2160&.15+2160 = 2484/.21 (by mole fraction of air to find total air)= 11828.6 mol air.

Extent of Raction:

......In...-.....Out...= Ev
CH4.....800.....-.....0...= -E1
C2H6...160.....-.....0...= -2E2
CO2......0...-.....nCO2... = -E1 - 2E2
H2O......0...-.....nH2O... = -2E1 - 3E2

where E is xi or extent of raction, and v is nu or the stoichiometric coefficient

From here on I just was lost, any ideas?

 

[KataKoniK]

I would approach this problem by first identifying the key variables and their corresponding equations. In this case, the key variables are the volume of gas (V), the number of moles of gas (n), the temperature (T), and the pressure (p). The equations that relate these variables are the ideal gas law (pV = nRT) and the stoichiometric equations for the combustion reaction.

Using the ideal gas law, we can calculate the number of moles of gas produced by the combustion reaction. The initial volume of gas (1000 ft^3/h) and temperature (60 degrees F) can be converted to standard conditions (760 mm Hg and 273 K) to make the calculations easier.

pV = nRT
(760 mm Hg)(1000 ft^3/h)(0.02832 m^3/1 ft^3)(1 atm/760 mm Hg)(273 K/60 F) = n(0.0821 L atm/mol K)(1 mol/22.4 L)
n = 118.7 mol

Next, we can use the stoichiometric equations to determine the number of moles of each gas produced. For simplicity, I will use the mole fractions given in the problem to calculate the moles of each gas produced.

nCH4 = (118.7 mol)(0.80) = 94.96 mol
nC2H6 = (118.7 mol)(0.16) = 18.99 mol
nO2 = (118.7 mol)(0.02) = 2.37 mol
nCO2 = (118.7 mol)(0.01) = 1.19 mol
nN2 = (118.7 mol)(0.01) = 1.19 mol

To find the total volume of gas produced, we can use the ideal gas law again, this time using the total number of moles and the given temperature and pressure.

pV = nRT
(760 mm Hg)(V)(0.02832 m^3/1 ft^3)(1 atm/760 mm Hg)(273 K/800 F) = (118.7 mol + 94.96 mol + 18.99 mol + 2.37 mol + 1.19 mol + 1.19 mol)(0.0821 L atm/mol K)(1 mol/