(chapter 4, A Reaction) Example:

.7044 mol/s Hexane and 9.287 mol/s Methane and 100 % excess air are fed into a reactor to produce only CO2 and H2O as products of the reaction. Determine all inlet and outlet flow rates, and the extents of reaction.

Flow Diagram:

Our Reactions

Here are the reactions that occur in our reactor.

C6H14 + O2 ® CO2 + H2O

CH4 + O2 ® CO2 + H2O

or, in getting the stoichimetry right, we have,

1C6H14 + 9.5O2 ® 6CO2 + 7H2O, and call this reaction #1

1CH4 + 2O2 ® 1CO2 + 2H2O, and call this reaction #2

And, the useful formulas are,

ni, leaving the reactor = ni,into the reactor + Sof all reactions (vi*x)

And, rearranging this to show that we can calculate x if we just know how many moles of any of the components that reacted (because we would already know vi for that species from the reaction),

x = ni, that reacted / |vi|

and finally, our percent excess formula,

%xs = (feed - stoich) / stoich * 100%, or = amount left / stoich

where, "stoich" referes to the number of moles that react.
Here now, we need not write out the first equation given for each component, because we can calculate the extents of reaction with the second equation, but we proceed to write out the first equation for each component, for illustrative purposes. Here, out referes to the outlet stream, and in referes to the respective inlet stream.

ndotout,Hex = .7044mol/s + (-1)*x1 + (0)*x2 = 0
ndotout,Me = 9.287mol/s + (0)*x1 + (-1)*x2 = 0
ndotout,O2 = ndotin,O2 + (-9.5)*x1 + (-2)*x2
ndotout,H2O = 0 + (+7)*x1 + (+2)*x2
ndotout,CO2 = 0 + (+6)*x1 + (+1)*x2 = 0
ndotout,N2 = ndotin,N2 + (0) + (0) = 0
The hexane and methane balances give both the extents of reaction, this is the first thing we would calculate, after this, we could sub in the extents of reaction into the other equations, thus, each of the above 6 equations readily give,

x1 = .7044 mol/s
x2 = 9.287 mol/s
ndotout,O2 = ndotin,O2 + -25.26 mol/s
ndotout,H2O = 23.50 mol/s
ndotout,CO2 = 13.51 mol/s
ndotout,N2 = ndotin,N2

Now, we can go no farther, and we find that we can still use the %xs relation, so, given air in 100% xs, this gives,

%xs air = [ndotO2 fed - ndotO2 stoich] / ndotO2 stoich * 100% = 100%

[ndotO2 fed - ndotO2 stoich] / ndotO2 stoich = 1

and since ndotO2 stoich = 9.5x1 + 2x2, and we have found the extents of reaction, thus we get ndotO2 stoich = 25.26 mol/s, now, if we plug this into our equation, we get ndotO2 fed = 50.52 mol/s.
Now, we can readily get the fed N2, and we get .79/.21 * 50.52 mol/s = 189.0 mol/s.
Now, going back to our O2 balance, with the feed O2 now determined for the %xs relation, we can now sub in this value to get the O2 in the outlet, to get, ndotout,O2 = 25.26 mol/s. Now, all flows have been determined as have both extents of reaction.

x1 = .7044 mol/s
x2 = 9.287 mol/s


All flowrates in mol/s.

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