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Daily output of NO from power plant.

  1. Oct 24, 2011 #1
    The problem statement, all variables and given/known data:

    Q.If a power plant burning 10,000 metric tons of coal per day with 10% excess air emits stack gas containing 100 ppm by volume of NO, what is the daily output of NO? (Assume coal is pure C).


    The attempt at a solution:

    A. C + O2 → CO2

    (10000 metric tons of C) / 12.0108 g/mol x (1/1) x (44.0096 g/mol) x (100/ 1000000) = 3.66 metric tons of NO

    Please can someone check this is correct for me. If incorrect please show me. Thanks,

    Ed
     
    Last edited: Oct 24, 2011
  2. jcsd
  3. Oct 24, 2011 #2
    - You forgot the 10% excess air
    - the atmosphere is only 21% oxigen (by volume)
    - you calculated 100 ppm by weight and not by volume

    start with calculating how many moles of C are present. You can then compute
    how many moles of NO are present without using the weight of anything, and then
    finally compute the weight of the NO.
     
  4. Oct 25, 2011 #3
    Moles of C present:

    10,000 / 12.0108 = 832.58 moles of coal. (do i need to convert 10,000 metric tons into grams? Other than that I have no clue what to do with the 100ppm by the volume. Please show me the working out to receive the correct answer then I can go through it and understand better. Thanks
     
  5. Oct 25, 2011 #4
    Yes, 1 mole of carbon is 12.0108 g,and not 12.0108 metric tons.

    the volume of 1 mole is the same for all the gases, so if 100 ppm by volume is NO, there's one 1 mole of NO, for every 10000 moles of gas coming out of the pipe.

    you need to add a volume of air that contains enough oxygen to react with the coal. Air is 21% oxygen by volume, so you need 100 moles of air to get 21 moles of oxygen and then you need 10% extra air.
     
  6. Oct 26, 2011 #5
    10,000 metric tons = 10 000 000 000 grams. (1.0 * 10^10)

    1 mole of C = 832.58 moles
    1 mole of NO = 832.58 moles

    So the equation would be:

    (1.0*10^10 grams of C) / 12.0108 g/mol x (1/1) x (44.0096 g/mol) x (100/ 10000) = 366416891.5 grams of NO?? Sorry I really don't understand this one
     
  7. Oct 27, 2011 #6
    don't try to write everyting in one equation, that way I don't understand it either.

    there are 10^10 / 12.0108 = 8.33 *10^8 moles of C present.

    these need 8.33 *10^8 moles of oxygen to burn

    1 mole of air contains 0.21 moles of oxygen so

    ........... moles of air contain 8.33 *10^8 moles of oxygen

    there's 10% extra air, so you need ............. moles

    100/1000000 parts of the exhaust is NO, so this is .......... moles, wich weigh ... g, so you get .......... kg of NO
     
  8. Oct 29, 2011 #7
    Yeah shouldn't of put it in that way. Ill put it in a clearer way in future.

    1 mole of air contains 0.21 moles of oxygen so

    3.9*10^9 moles of air contain 8.33 *10^8 moles of oxygen

    there's 10% extra air, so you need 3.57*10^9 moles

    100/1000000 parts of the exhaust is NO, so this is 3.57*10^5 moles,

    which weigh 10713570g, so you get 10713.570 kg of NO.
     
  9. Nov 1, 2011 #8
    Ignore post above (wrong)

    1 mole of air contains 0.21 moles of oxygen so

    3.9*10^9 moles of air contain 8.19*10^8 moles of oxygen

    there's 10% extra air, so you need 1.209*10^9 moles

    100/1000000 parts of the exhaust is NO, so this is 1.209*10^5 moles,

    which weigh 3.638*10^6g, so you get 3628kg of NO. Willem2 please check :)
     
  10. Nov 1, 2011 #9
    Assume air is 21% O2 and is at STP (273 K, 1.00 atm)

    From the stoichiometry of the reaction C + 02 --> CO2 the mass of 02 needed is given by:

    Mass O2 = 10,000 tonne x (32 tonne O2/ 12 tonne C) 2.67 x 10^4 tonne O2

    Stoichiometric mass air 2.67 x 10^4 tonne/0.21 = 1.27 x 10^5 tonne air 1.27 x 10^11 g air, To account for 10% excess air:

    Mass air = 1.27 x 10^5 tonne x 1.10 = 1.40 x 10^5 tonne air 1.40 x 10^11 g air

    Using 1.29 g/L for the density of air, its volume is given by

    V 1.40 X 10^11 g/1.29 g/L = 1.09 x 10^11 L

    Mol air = 1.09 x 10^11 L/(22.4 L/mol) = 4.84 x 10^9 mol air

    At 100 ppm NO, the mol fraction of NO is 1.00 x 10^-4

    Mol NO = 1.00 x 10^-4 mol NO/mol air x 4.84 x 10^9 mol air = 4.84 x 10^5 mol NO

    Mass NO = 4.84 x 10^5 mol NO x 30.0 g NO/mol NO = 1.45 x 10^7 g NO

    The daily output of NO is 14.5 tonne/day.

    This seems an easier way.

    Could anyone check please.
     
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