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Reducing Smoke

  1. May 28, 2005 #1


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    Is there away tio reduce the smoke produced in a [combustion] reaction?
    Theoretically, there is no smoke produced at all (only water and carbon dioxide), but because of the impurities, imcomplete combustion, moisture, obviously, smoke is a common product of burning something.
    How can I minimilize smoke from being produced?
    Some stuff I use makes alot more smoke for instance than others,
    for example, Potassium Nitrate makes alot of white (smelly) smoke, where are buring alot of oily rags would produce black smoke, and by contrast, combusting hydrogen and oxygen to produce water makes little if any smoke.
  2. jcsd
  3. May 28, 2005 #2
    Smoke (in gunpowder) is produced by incomplete combustion (hence solid produces). In gunpowder, potassium nitrate the main source of oxygen and is used to make more gases and less solid products (reducing the smoke levels). However it is still quiet smokey (as you have found).

    Simply, the more gases you can produce, the less smoke there will be. To make combustion more complete, trying adding a compound that has high oxygen to improve the combustion. Smoke may be produced still but it should be less.

    As you can tell from the reaction of hydrogen and oxygen, if you removed some of the oxygen there would be spare hydrogen. If the oxidising compound was potassium nitrate then the smoke levels would increase because the oxygen has been removed and the potassium and nitrogen are left (solid product: potassium or potassium oxide). As you can see, the more oxygen to other elements ration you have the less smoke you will get.

    This probably did not help at all and I am likely to be corrected by a superior body but still....... :smile:

    The Bob (2004 ©)
  4. May 28, 2005 #3


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    I am already using compounds that will supply lots of exygen to the reaction (no to mention anything it may grab from the air), and I still get more smoke that I'd like.
    Just for instance, a KNO3 and sugar combustion,
    5C12H22O11 + 48KNO3 -> 36CO2(g) + 55H2O(g) + 24N2(g) + 24K2CO3,
    this reaction produces an astonishing 105 moles of gas, and yet, by comparison to a very similar reaction (chlorate and sugar), it produces much more smoke,
    C12H22O11 + 8KClO3 -> 12CO2(g) + 11H2O(g) + 8KCl,
    which only produces 23 moles of gas and has a similar ratio of gas to "other" stuff, than the previous nitrate reaction.
    then again, the above is based on my own experience using my own chlorate and nitrate, it is possible that the nitrate has other impurities that I dont know about in it that is causing that extra smoke.
    Last edited: May 28, 2005
  5. May 30, 2005 #4
    Just quickly:
    5C12H22O11 (s) + 48KNO3 (s) -> 36CO2 (g) + 55H2O (g) + 24N2 (g) + 24K2CO3 does not balance. Assuming this reaction is in air then it could be:
    C12H22O11 (s) + 12KNO3 (s) + 13.5 O2 (g) -> 6CO2 (g) + 11H2O (g) + 6N2 (g) + 6K2CO3 (I couldn't balance it with what we had).

    Right, so, maybe you have found that more gases (with solid products) mean that the solids are dispersed more.

    It might be possible to remove the potassium nitrate and create no solid products. The combustion will be incomplete but most are.

    The Bob (2004 ©)
  6. May 30, 2005 #5


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    ...for 5 moles of sugar, or 21 moles of gas per mole of sugar. Not terrible compared to 23 (even if correct).

    You need to better define what you mean by smoke and exactly what you want to minimize. The number of moles of water and carbon dioxide are fixed by the combusting substance, so there is no way of changing that. Usually, however, neither of these gases is considered smoke. Smoke is primarily a colloidal suspension of unburnt carbon particles carried by the evolved gases.

    If you want to minimize the unburnt carbon, one thing that helps is having the reaction be close to equilibrium (ie : ensure a slow reaction rate).
  7. Nov 9, 2011 #6
    OK, I'm confused... It seems to me 5C12H22O11 + 48KNO3 -> 36CO2 + 55H2O + 24N2 + 24K2CO3 does balance, and there are 115 not 105 mols of gas produced for 5 mols of sugar. What am I missing here?
  8. Nov 9, 2011 #7


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    Nothing on the chemical side. Reaction is correctly balanced and it produces 115 moles of gas.

    The only thing you missed is that this thread is over 6 years old.
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