.7044 mol/s Hexane and 9.287 mol/s Methane and 100 % excess air are fed into a reactor to produce only CO
Our Reactions
Here are the reactions that occur in our reactor. _{6}H_{14} + O_{2} ® CO_{2} + H_{2}OCH _{4} + O_{2} ® CO_{2} + H_{2}O_{6}H_{14} + 9.5O_{2} ® 6CO_{2} + 7H_{2}O, and call this reaction #11CH _{4} + 2O_{2} ® 1CO_{2} + 2H_{2}O, and call this reaction #2And, the useful formulas are, _{i, leaving the reactor} = n_{i,into the reactor} + S_{of all reactions} (v_{i}*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 v _{i} for that species from the reaction), _{i, that reacted} / |v_{i}|and finally, our percent excess formula, 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. _{out,Hex} = .7044mol/s + (-1)*x_{1} + (0)*x_{2} = 0ndot _{out,Me} = 9.287mol/s + (0)*x_{1} + (-1)*x_{2} = 0ndot _{out,O2} = ndot_{in,O2} + (-9.5)*x_{1} + (-2)*x_{2}ndot _{out,H2O} = 0 + (+7)*x_{1} + (+2)*x_{2}ndot _{out,CO2} = 0 + (+6)*x_{1} + (+1)*x_{2} = 0ndot _{out,N2} = ndot_{in,N2} + (0) + (0) = 0_{1} = .7044 mol/s x _{2} = 9.287 mol/sndot _{out,O2} = ndot_{in,O2} + -25.26 mol/sndot _{out,H2O} = 23.50 mol/sndot _{out,CO2} = 13.51 mol/sndot _{out,N2} = ndot_{in,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, _{O2 fed} - ndot_{O2 stoich}] / ndot_{O2 stoich} * 100% = 100%or, _{O2 fed} - ndot_{O2 stoich}] / ndot_{O2 stoich} = 1and since ndot _{O2 stoich} = 9.5x_{1} + 2x_{2}, and we have found the extents of reaction, thus we get ndot_{O2 stoich} = 25.26 mol/s, now, if we plug this into our equation, we get ndot_{O2 fed} = 50.52 mol/s.Now, we can readily get the fed N _{2}, and we get .79/.21 * 50.52 mol/s = 189.0 mol/s. Now, going back to our O _{2} balance, with the feed O_{2} now determined for the %xs relation, we can now sub in this value to get the O_{2} in the outlet, to get, ndot_{out,O2} = 25.26 mol/s. Now, all flows have been determined as have both extents of reaction.
_{1} = .7044 mol/s x _{2} = 9.287 mol/s
All flowrates in mol/s. |

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