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    • Thread Starter

    For the half equation of:

    NO3- ---> N2

    Is this correct:

    2NO3- + 6H+ + 6e- ---> N2 + 3H20

    If not why?

    2NO3 + 10e + 12H+ → N2 + 6H2O
    • Thread Starter

    (Original post by Holby_fanatic)
    2NO3 + 10e + 12H+ → N2 + 6H2O
    Why 10 rather than 12 electrons?

    Also Nitric acid is produced by these 3 reactions

    4NH3 + 5O2 ---> 4NO + 6H2O

    2NO + O2 ---> 2NO2

    3NO2 + H2O ---> 2HNO3 + NO

    Starting with 1000kg ammonia, whats the maximum mass of HNO3?

    I'm struggling here :

    what I did was

    1000000/17 = mass of nh3 in grams divided by mr nh3 to get moles of nh3

    (1000000/17) which is 4 moles by the reaction, and divide that by 2 to get

    (1000000/34).-2 moles by the reaction

    Then multiply by Mr of HNO3, 63

    (1000000/34 x 63)/1000 = 1852.94 kg

    But the answer is 2471 kg. Where am I going wrong? Any help greatly appreciated

    If 12 electrons were used, the equation would look something like this
    2NO3 + 12e + 12H+ --> N2 + 6H2O +2e-
    There would be two spare electrons as all products would already have a neutral charge.

    For the second part of your question, I must admit I was struggling. I have two methods, the first one is completely wrong because I ended up with the same answer as you.

    METHOD 1
    4NH3 + 5O2 ---> 4NO + 6H2O
    As there is a 1:1 ratio between NH3 and NO, 1000kg of NO could theoretically be made.
    Moles of NH3 = 1000,000g / 17g mol-1
    = 58823.5 mol
    This is for 4 moles of NH3 and so 1 mole of NH3 = 58823.5 / 4
    =14706 mol

    2NO + O2 ---> 2NO2
    Half of the moles NO produced in the first reaction is used in the second reaction. In this reaction, there is again a 1:1 ratio between NO and NO2.
    Moles of NO used = 58823.5 / 2
    = 29411.8 mol
    Therefore, mass of NO2 produced = 29411.8 mol

    3NO2 + H2O ---> 2HNO3 + NO
    The moles of NO2 used is 1.5 times that of the moles produced in the second reaction. There is a 3:2 ratio between NO2 and HNO3
    Moles of NO2 used = 29411.8 mol x 1.5
    = 44117.7 mol
    Moles of HNO3 produced = 2/3 mass of NO2 used
    = 44117.7 x 2/3
    = 29411.8 mol

    Mass of HNO3 produced = number of moles x MR
    = 29411.8 mol x 63g mol-1
    = 1853kg

    METHOD 2
    Say 1000kg of NH3 results in 1000mol of NO
    1000mol of NH3 produces 1000mol of NO
    As half of the number of moles of NO produced is needed for the second reaction half of the mass of NO is needed. 500mol of NO is used.
    500mol of NO produces 500mol of NO2
    As 1.5 times the number of moles of NO2 produced is needed for the third reaction 1.5 times the mass of NO2 is needed. 750mol of NO2 is used.
    750mol of NO2 produces 375mol of HNO3. (This is because 3 moles of NO2 is needed while only 2 moles of HNO3 is produced 2/3 x 750mol = 375mol)

    Mass of HNO3 = moles x MR
    = 375mol x 63g mol-1
    = 2362.5kg

    Do you need these answers urgently or do you have some time? I'll keep trying to work it out until I get the right answer. (This is much more fun than writing an essay on polymers).
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