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Chemistry: gases, exercise

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello!
    I would be grateful for your help on finding my mistake - somehow my answers are
    wrong compared to the answer from the book.
    Here is the exercise from a paragraph on gases (cumulative gas law, basics):

    Consider the reaction for the synthesis of nitric acid:
    3NO2 + H2O = 2HNO3 + NO
    (a) If 12.8 L of NO2 measured at STP, is allowed
    to react with 14.9 g of water, find the limiting
    reagent and the theoretical yield of HNO3 in
    grams.

    Here is my solution:

    1 mole of gas at STP = 22.4 L; 12.8 L gives 0.5714 moles of NO2
    1 mole of H2O = 18.016 g; 14.9 g = 0.827 moles of H2O

    The limiting reagent is NO2, because
    3 moles NO2 + 2 moles of H2O
    0.5714 moles NO2 requires only 0.381 moles of H2O, but we have 0.827 moles of H2O

    1 mole of NO2 = 46.007 g
    0.5714 moles of NO2 = 26.288 g
    Theoretical yield of HNO3:
    3 moles of NO2 = 2 moles of HNO3
    0.5714 moles of NO2 = 0.3809 moles of HNO3

    1 mole of HNO3 = 63.008
    0.3809 moles of HNO3 = 24.001 g

    Thank you!
     
  2. jcsd
  3. Mar 6, 2016 #2

    mfb

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    2016 Award

    Staff: Mentor

    Why 2 moles of H2O?
    The conclusion is still right.

    Looks right.
    I don't think you have to work with decimal numbers of molar masses, they are very close to integers for the involved elements.
     
  4. Mar 6, 2016 #3

    Borek

    User Avatar

    Staff: Mentor

    There is the mistake mfb already pointed out, but it doesn't influence the final answer.

    What is the answer given in the book?
     
  5. Mar 6, 2016 #4
    Thank you! Sorry, yes, only one mole of H2O.
     
  6. Mar 6, 2016 #5
    I think the problem was in the question - there is some typo in the book, because exercise is asking about theoretical yield of HNO3,
    while the book gives answer on grams of NO2
     
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