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Multiply by the number of moles of that substance

  1. Mar 21, 2006 #1
    How do I figure out how many moles to mutiply by?

    It says "multiply by the number of moles of that substance"

    How do I figure that out?

    Here's where I am:

    H2O(l)= - 285.830 kJ/mol x ___ mol

    NH3(g)= -46.11 kj/mol x ___ mol


    And what does the L or G mean for the equation?:grumpy:
  2. jcsd
  3. Mar 21, 2006 #2
    Is this an exercise?
    If it is, they must say the number of moles of H2O and the number of moles of NH3.

    l and g are the states of water and ammonia and they stand for liquid and gaseous, respectively. In this situation, I think they are not important, but I don't understand well this exercise.
  4. Mar 21, 2006 #3
    This is the question:

    Balance the equation and Calculate the enthalpy change for the following reaction:

    NH3(g) + O2(g) --> N2(g) + H2O(l)

    NH3(g)= -46.11 kj/mol

    O2(g) = 0 kJ/mol

    N2(g) = 0 kJ/mol

    H20(l)= - 285.830 kJ/mol

    I know that to begin, I need to multiply the numbers above by the amount of moles in the substances. I think. LOL
  5. Mar 21, 2006 #4
    First, we balance the equation:

    2NH3(g) + 1.5O2(g) --> N2(g) + 3H2O(l)

    Enthalpy change = enthalpy of products - enthalpy of reactants
    Enthalpy change = 2 mol x -46.11 kj/mol - 3 mol x - 285.830 kJ/mol
    Enthalpy change = -92.2 + 857.49
    Enthalpy change = 765.29 kJ

    Note: I haven't learnt yet to resolve enthalpy change problems in school like this, but is the only way I see to resolve it. I seriously advise you to wait for a confirmation or correction of this resolution. I just tried to help.
    Last edited by a moderator: Mar 21, 2006
  6. Mar 21, 2006 #5
    Looking over my notes, it looks like you hit it right on! I'm still a *little* fuzzy but I can't thank you enough for helping me out. I was gonna have to submit this with no idea what I was doing.. :) Any help at all is better than no help. Thankyou SO much for your help.
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