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Topic: Assignment about a gas mixture that is burned  (Read 4384 times)

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Offline MNIO

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Re: Assignment about a gas mixture that is burned
« Reply #30 on: May 14, 2020, 03:50:22 PM »
Borek,

The issues with the problem are (1) the individual values for x, y, z are unknown and (2) to complicate the problem, O2 is in XS.

We could certainly write.
  x CO + 1/2x O2 ---> x CO2
  y CH4 + 2y O2 ---> y CO2 + 2y H2O
  z N2 ---> z N2
add then
  x CO + y CH4 + z N2 + (1/2*x + 2*y) O2 --> (x+y) CO2 + 2y H2O + z N2

But in this case, that might lead to the mistake of assuming 1/2*x + 2*y = 150.  So to correct for this, we need to assume some incremental O2 that remains unreacted.  Let's call it "w".  So we could write
  x CO + y CH4 + z N2 + (1/2*x + 2*y + w) O2 --> (x+y) CO2 + 2y H2O + z N2 + w O2

then
  x+y = 220 - 80 = 140
  x+y+z = 200
  z = 200 - 140 = 60
  w = 80 - 60 = 20
then
  1/2*x + 2y + 20 = 150
and we get
  1/2 x + 2y = 130
  x + y = 140
and the end game is the same. 
  100 CO + 40 CH4 + 60 N2 + 150 O2 --> 140 CO2 + 80 H2O + 60 N2 + 20 O2

The question is... would Mathjr catch that "w" needs to be added.  And would Mathjr know to use x, y and z as coefficients of CO, CH4 and N2. And of course there's the cm3 instead of moles and the reaction of NaOH with CO2. 

*******
Anyway, the way I originally showed was to put the big picture together first, but if this works better, so be it. But it is still an exercise in balancing an equation with a twist (that we've all worked in past lives  :) )
« Last Edit: May 14, 2020, 04:34:40 PM by MNIO »

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