1. Hey Guest! Check out the 3 MCAT Study Plan Options listed in the 'stickies' area at the top of the forums (BoomBoom, SN2ed, and MCATJelly). Let us know which you like best.

    Also, we now offer a MCAT Test-Prep Exhibitions Forum where you can ask questions directly from the test-prep services.
    Dismiss Notice
  2. Download free Tapatalk for iPhone or Tapatalk for Android for your phone and follow the SDN forums with push notifications.
    Dismiss Notice
3 MCAT Study Options Tapatalk
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

Dismiss Notice
Hey Texans—join us for a DFW meetup! Click here to learn more.

stoichiometry problem?

Discussion in 'MCAT Discussions' started by PdiM, Jul 26, 2002.

Previous Thread Next Thread
Loading...
  1. PdiM

    PdiM Junior Member
    7+ Year Member

    Joined:
    Jul 20, 2002
    Messages:
    11
    Likes Received:
    0
    Hey,

    I had a problem with the following question from a Kaplan simulated exam. Maybe somebody would be able to help me out:

    given
    HNO3 --> HNO2 --> H2N2O2 --> N2O(g) --> N2 (g)

    If all the oxygen in the nitric acid is converted to water, how many additional equivalents of acid will be consumed during the production of 5M of nitrogen?

    a) 20 b) 30 c) 40 d) 50

    i've been stratching my head with this one, but can't seem to figure it out. I havent taken a look at kaplan's explanation yet. any takers??
     
    #1 PdiM, Jul 26, 2002
    2. Note: SDN Members do not see this ad.

  2. Mudd

    Mudd Charlatan & Trouble Maker
    7+ Year Member

    Joined:
    Jun 5, 2002
    Messages:
    376
    Likes Received:
    2
    My assumption is that the arrows represent intermediates, which don't matter much in a stoichometry question. Assuming the overall reaction is:

    HNO3 ---> H2O + N2

    Then the balanced version, excluding hydrogens, is:

    2 HNO3 ---> 6 H2O + 1 N2

    There is a shortage of 10 H atoms on the reactant side of the equation. To generate 5 moles of N2 would require an additional 50 moles of protons from some source.

    Something is wrong with this problem however, because from HNO3 to N2 is reduction, and there is no oxidation half-reaction shown. Is there more to the question?
     
    #2 Mudd, Jul 26, 2002
  3. PdiM

    PdiM Junior Member
    7+ Year Member

    Joined:
    Jul 20, 2002
    Messages:
    11
    Likes Received:
    0
    the answer IS 50 :)

    wow i was approaching that question from a totally wrong perspective - i thought when they were asking for "acid equivalents" they meant additional HNO3.

    This was all that was stated in the question. Does it HAVE to be accompanied with the oxidation half-reaction? they couldnt just show the reduction half-reaction?
     
    #3 PdiM, Jul 26, 2002
  4. Tweetie_bird

    Tweetie_bird
    7+ Year Member

    Joined:
    Nov 20, 2001
    Messages:
    2,193
    Likes Received:
    2
    aaww you little teaser
    Didn't even give the answer!

    Anyway, is it "B"....30 extra acid equivalents? Lemme know before the curiosity kills me. :D
     
    #4 Tweetie_bird, Jul 26, 2002
  5. PdiM

    PdiM Junior Member
    7+ Year Member

    Joined:
    Jul 20, 2002
    Messages:
    11
    Likes Received:
    0
    hehe i thought it would drive people to actually work out the problem if i didnt give the answer ;)

    its d) 50
     
    #5 PdiM, Jul 26, 2002
  6. Mudd

    Mudd Charlatan & Trouble Maker
    7+ Year Member

    Joined:
    Jun 5, 2002
    Messages:
    376
    Likes Received:
    2
    You are correct, there is no need to specify an oxidation half-reaction, but to balance the reaction, they need 10 e- on the reactant side for every N2 that is formed. This would have allowed you to balance the reaction by charge, rather than stoichiomtery. Seeing 10 e- would have made it more clear that 10 H+ were required, or that 10 HNO3 would provide the 10 H and yield 10 nitrate anions in the product of the reduction half-reaction.
     
    #6 Mudd, Jul 26, 2002
Previous Thread Next Thread
Loading...
(You must log in or sign up to post here.)

Share This Page