1. The SDN iPhone App is back and free through November! Get it today and please post a review on the App Store!
    Dismiss Notice
  2. Dismiss Notice
Dismiss Notice

Interview Feedback: Visit Interview Feedback to view and submit interview information.

Interviewing Masterclass: Free masterclass on interviewing from SDN and Medical College of Georgia

limiting reagent

Discussion in 'MCAT Study Question Q&A' started by Oh_Gee, Sep 7, 2014.

  1. Oh_Gee

    5+ Year Member

    Joined:
    Nov 14, 2013
    Messages:
    1,602
    Likes Received:
    885
    Status:
    Medical Student
    will the reactant with less moles always be the LR? do the coefficients of the reactants matter?
     
  2. Note: SDN Members do not see this ad.

  3. Cawolf

    5+ Year Member

    Joined:
    Feb 26, 2013
    Messages:
    3,365
    Likes Received:
    2,094
    Status:
    Medical Student
    The substrate that is used up first.

    So it depend on coefficients.

    For example, 10A + B -----> A10B

    If there are 10 moles of A and 5 moles of B, the LR will be A even though it has more moles.

    The reaction will use 10 moles of A and only 1 of B, leaving 4 moles of B.
     
  4. Oh_Gee

    5+ Year Member

    Joined:
    Nov 14, 2013
    Messages:
    1,602
    Likes Received:
    885
    Status:
    Medical Student
    so if you were given grams of A and B and you calculated that you have 1 mol A and 3 mols B, A would still be the LR? and .1 mols of A10B will be made?
     
  5. Cawolf

    5+ Year Member

    Joined:
    Feb 26, 2013
    Messages:
    3,365
    Likes Received:
    2,094
    Status:
    Medical Student
    Yup exactly.

    You can calculate your moles and then use dimensional analysis to check for the limiting reagent.

    Given 1 mol A and 3 mol B

    1 mol A * (1 mol A10B/10 mol A) = .1 mol A10B

    3 mol B * (1 mol A10B/1 mol B) = 3 mol B

    You can only go as far as your weakest link!

    So then we can say that a reaction creating .1 mol A10B * (1 mol B/1 mol A10B) = .1 mol B will be used

    So we started with 3 mol B and used .1 mol B = 2.9 mol B remaining.
     

Share This Page