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How do I know which one is the limiting reactant?

  1. Sep 22, 2013 #1
    1.
    I am doing step number 9 to determine the limiting reactant for this lab.

    http://www.lahc.edu/classes/chemistry/arias/Exp 4 - Limiting RF11.pdf

    Basically I have two vials labeled A and B filled with filtrate that is yellow.

    For vial A, I added 10 drops of 0.2 M K2CrO4, but nothing happened. The color stayed the same.

    For vial B, I added 10 drops of 0.1 M AgNO3, and the whole thing turned red.


    I am thinking since AgNO3 made the filtrate turn red then it is the limiting reactant, but I am not sure.

    Can somebody help me?



    Furthermore, when I do the math the grams of AgNO3 is always bigger than K2CrO4.

    I am wondering if it is possible for the limiting reactant to weigh more (in grams) than the excess reactant?






    2. Relevant equations



    3.
    Finding the weight of the limiting and excess reactant:

    The precipitate (Ag2CrO4) that forms weights 0.193 g and the moles are 5.82 x 10^-4 mol.



    Finding for mass for AgNO3:

    (5.82 x 10^-4 mol Ag2CrO4) (2 mol AgNO3 / 1 mol Ag2CrO4) (169.872 g AgNO3 / 1 mol AgNO3)

    = 0.198 g AgNO3


    Finding for mass for K2CrO4:

    (5.82 x 10^-4 mol Ag2CrO4) (1 mol K2CrO4 / 1 mol Ag2CrO4) (194.188 g K2CrO4 / 1 mol K2CrO4)

    = 0.113 g K2CrO4



    AgNO3 always comes out bigger than K2CrO4. So is it possible for AgNO3 to be the limiting reactant?

     
  2. jcsd
  3. Sep 22, 2013 #2

    Simon Bridge

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    You have to use stochiometry to figure out which is the limiting reactant - it's the one that got completely used up.
     
  4. Sep 22, 2013 #3
    I used stoichmetery beginning with the moles of the precipitate, and found the grams of both reactant. AgNO3 will always be larger than K2CrO4
     
  5. Sep 22, 2013 #4

    Simon Bridge

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    This statement does not make a lot of sense. When the reaction has stopped, one of the reactants should have been used up completely.
    i.e. there should be zero grams of one of them. So what did you just calculate - precisely?

    I think you have another step to do. The thought process is something like this:
    How much of each reactant did you mix together?
    How many of each atom does that give you?
    What is the maximum number of the product molecule this can make?
    Which reactant decides what this last number is?

    Note:
    You found the amount of each reactant. ("grams" is a unit of mass - the amount of substance)
     
    Last edited: Sep 22, 2013
  6. Sep 23, 2013 #5
    On step 2 of the lab work of day 2 it said use stoichiometry to find the mass which I did and got:

    0.198 g AgNO3

    Then I used stoichiometry to find the mass of the other reactant which is:

    0.113 g K2CrO4


    Would these be the starting amounts I mixed together?



    Note:
    I use the value of the product (Ag2CrO4 = 0.193 g) to find the mass of AgNO3.
     
  7. Sep 23, 2013 #6

    Simon Bridge

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    Where did you get the numbers from? Did you get the from the starting amounts?
    i.e. is that how much of each reactant is in the 10 drops?

    If you had worked out final amounts, one of them would be zero right?
    So they are probably starting amounts - the thing is: you should know - you are the one who did the math.

    The reactants are the two things in the vial together.
    You are adding K2CrO4 to AgNO3
    You get Ag2CrO4 (and some other stuff.)

    How many of each reactant goes to make one molecule of the product?
    If there were 100x more K2CrO4 than AgNO3, then the AgNO3 would be the limiting reactant right?


    What is the quantity that tells you the number of each molecule you added?
     
  8. Sep 23, 2013 #7

    To get the starting amount of the precipitate I had to go through the heating and boiling the substances portion then I waited to filter out the precipitate after it was done boiling.

    Then, I weighted the precipitate which was Ag2CrO4 = 0.193 g

    and with that value I had to find the mass of the limiting reactant. Which I would choose which is the limiting reactant by the experiment oberservation:

    """For vial A, I added 10 drops of 0.2 M K2CrO4, but nothing happened. The color stayed the same.

    For vial B, I added 10 drops of 0.1 M AgNO3, and the whole thing turned red.""""


    The thing is if I do the math to see how much of the product (KNO3) formed I would get the same value using the starting amount of the reactants. I thought one of the values is supposed to be lower than the other.
     
  9. Sep 23, 2013 #8

    Simon Bridge

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    Well yeah - one of the reactants is used up. Though in some cases the concentration just drps so low the recation no longer happens.

    You have to use your understanding of the experiment to figure out which one that was.

    The difference between the two vials is important of course. Describe that step more carefully.
    All this discussion is aimed at getting you to think about how the experiment was supposed to work.
     
  10. Sep 23, 2013 #9
    I think that's why the math is coming out the why it is and why the amount of product that form is the same value for either starting amount of reactant I use.

    So, the only way in this experiment to figure out which is the limiting reactant is by the experiment observation.

    """For vial A, I added 10 drops of 0.2 M K2CrO4, but nothing happened. The color stayed the same.

    For vial B, I added 10 drops of 0.1 M AgNO3, and the whole thing turned red.""""

    Since, the 0.1 M AgNO3 reacted it must have been limiting. I'm thinking that the K2CrO4 did not react because it was in excess.
     
  11. Sep 23, 2013 #10

    Simon Bridge

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    Because your product is a red precipitate?
    Ergo, no red no reaction (or very little reaction).

    Excess or not occurs after the reaction not before.
    What is stopping the reaction from happening in the second case?
    http://chemistry.about.com/od/examp...-Limiting-Reactant-Of-A-Chemical-Reaction.htm
     
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