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Limiting Reagent

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data

    When gaseous ammonia is passed over solid copper (II) oxide at high temperatures, nitrogen gas is formed.

    2NH3 + 3CuO → N2 + 3Cu + 3 H2O

    What is the limiting reagent when 34 grams of ammonia form 26 grams of nitrogen in a reaction that runs to completion?


    2. Relevant equations



    3. The attempt at a solution

    Using stoichiometry to find the solution in terms of N2, I get:

    34 g NH3 x 1 mol NH3/17 g NH3 x 1 mol N2/2 mol NH3

    = 1 mol N2

    Then,

    3 mol Cuo x 1 mol N2/3 mol CuO

    = 1 mol N2

    What am I doing wrong? I have tried this problem 7 times.
     
  2. jcsd
  3. Jul 9, 2014 #2

    Borek

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    Staff: Mentor

    You are not told how many moles of CuO were present, so you can't use them directly in your calculations.

    If there were 26 grams of nitrogen produced, did all ammonia react?
     
  4. Jul 9, 2014 #3
    That's what I am trying to figure out. However, I can find the grams of CuO produced by starting with 34 grams of ammonia, converting it to moles, then using the mole-mole ratio. I get 240 grams of CuO.

    The problem is determining the LR from here. When I use the grams of Cuo as well as the grams of ammonia (in terms of nitrogen gas), I keep getting 1 mole for each.

    How is this problem different from other basic stoichiometry problems? Is it because normally we are given the amounts of both reactants which makes it easier to determine the LR?
     
  5. Jul 9, 2014 #4

    Borek

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    Staff: Mentor

    That's not how much CuO was present. That's how much would be necessary to react with all ammonia. This is not necessarily amount of CuO present.

    You calculated how much would be needed to react with ammonia, no wonder it fits amount of ammonia.

    It is not different - or, if you see at as different, that's because you were not trained to see the whole picture, just part of it.

    Let me put the question differently: if there were 26 grams of nitrogen produced, how many grams of ammonia reacted?
     
  6. Jul 9, 2014 #5
    So, I have to work backwards on this one?

    For example, begin with 26 grams of nitrogen gas and convert to moles of ammonia = 1.86 mol ammonia

    Then from 26 grams of nitrogen gas, find the moles of CuO = 2.78 mol CuO

    Last, compare each with the mole ratios = .92 mol nitrogen (in terms of CuO); .93 mol nitrogen (in terms of ammonia).

    Am I on the right track? Sorry for the flow-chart like description.

    Thanks.
     
  7. Jul 9, 2014 #6

    Borek

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    Staff: Mentor

    Yes.

    And what is mass of 1.86 moles of ammonia?

    You are doing everything to not answer the simple question that I asked, but to mix it with irrelevant data.
     
  8. Jul 9, 2014 #7
    Limiting Reagents

    1.86 moles of ammonia = 31.62 grams of ammonia.
     
  9. Jul 10, 2014 #8
    Limiting Reagents

    Okay, I am still confused about this problem.

    2 NH3+ 3 CuO → N2+3Cu + 3 H2O

    Here is what I did:

    26 g N2 x 1 mol N2/28 g N2 x 3 mol Cuo/1 mol N2 =

    2.78 mol CuO

    26 g N2 x 1 mol N2/28 g N2 x 2 mol NH3/ 1 mol N2 = 1.85 mol NH3

    2.78 mol CuO x 80 g CuO/1 mol CuO = 222.4 g CuO

    1.85 mol NH3 x 17 g NH3/ 1 mol NH3 = 31.45 g NH3

    Knowing that the amount of NH3 used up is less than the provided amount in the problem and also considering that the reaction requires 222.4 mol CuO, am I to assume that CuO is limiting because the ammonia is 31.45 g?
     
  10. Jul 10, 2014 #9

    Borek

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    Staff: Mentor

    And that's all you need to know - by definition, limiting reagent is the one used to the end. Ammonia was not used to the end, so it was in excess. If it was in excess, CuO was the limiting reagent.

    Everything else you did was just confusing you. Especially, you don't have to calculate amount of CuO once you know ammonia was in excess.
     
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