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Homework Help: Chemical Reaction Problem

  1. Mar 12, 2010 #1

    Char. Limit

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    1. The problem statement, all variables and given/known data
    Give the formula to show the reactants and products for the following chemical reaction. The reaction occurs in aqueous solution unless otherwise indicated. Represent substances in solution as ions if the substance is extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. In all cases a reaction occurs. You need not balance (Poster's note: I'd like to try to balance anyway, but I'll do it on my own.)

    a. Solid calcium carbonate is strongly heated.
    b. A piece of nickel metal is immersed in a solution of cupric sulfate.
    c. Equal volumes of equimolar solutions of disodium hydrogen phosphate and hydrochloric acid are mixed.
    d. Chlorine gas is bubbled into a solution of sodium bromide.
    e. Ammonia gas is bubbled into a solution of ethanoic acid.
    f. Solid ammonium carbonate is added to a saturated solution of barium hydroxide.
    g. Drops of liquid dinitrogen trioxide are added to distilled water.
    h. Solutions of potassium permanganate and sodium oxalate are mixed.


    2. Relevant equations

    I don't know, but there may be some here that I don't know about.

    3. The attempt at a solution

    a. CaCO3(s) --> CaO(s) + CO2(aq)
    b. Ni(s) + Cu2+(aq) --> Ni2+(aq) + Cu(s)
    c. 2 Na+(aq) + 2 H+(aq) + PO43-(aq) + Cl-(aq) -->
    d. Cl2(g) + 2 Br-(aq) --> Br2(g) + 2 Cl-(aq)
    e. NH3(g) + H+(aq) --> NH4+(aq)
    f. (NH4)2CO3(s) + Ba(OH)2(aq) -->
    g. N2O3(l) + H2O(l) --> 2 H+(aq) + 2 NO2-(aq)
    h. K+(aq) + 2 Na+(aq) + MnO4-(aq) + C2O42-(aq) -->

    I'm slightly stuck on c and f (I think I can do them if I really concentrate), but I'm lost on h. I got a hint that it involves a redox reaction. Any tips on the three? And are the other five right?

    EDIT: I believe I've done f...

    (NH4)2CO3(s) + Ba2+(aq) --> 2 NH4+(aq) + BaCO3(s)
     
    Last edited: Mar 12, 2010
  2. jcsd
  3. Mar 12, 2010 #2

    Borek

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    c. HPO42- is a base.

    e. Slightly tricky - you are right about product being ammonium acetate, but I am not sure how to write reaction equation. I believe some teachers may prefer HA + NH3.

    f. You are close, but not there yet. What about OH-?

    h. Yes, it is definitely redox. If C2O42= looks to you like 2 molecules of carbon dioxide kept together by two electrons - you are right :wink:

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  4. Mar 12, 2010 #3

    Char. Limit

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    On f, the hydroxide ion is aqueous in both solutions, and a spectator ion thus, so I omitted it.

    On e, I just thought the same thing as in f: the spectator ion (acetate) is omitted.

    On h, I'm still a bit confused. We haven't learned too much about redox reactions yet. Could you give me a few general ideas?

    On c... so if biphosphate is a base, is disodium hydrogen phosphate a base as well? And then the reaction could be written...

    Na2HPO4 + 2 HCl --> H3PO4 + 2 NaCl...

    But no. An acid reacts with a base to form a salt and water...
     
  5. Mar 12, 2010 #4

    Borek

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    What other ions are in the solution?

    And that's not bad approach. Trick is, acetic acid is a weak one, so it is not fully dissociated. Whether it dissociates first and H+ reacts with base, or whether it is base reacting directly with acetic acid is a matter of approach.

    Start here: balancing redox reactions and read two next pages as well.

    Not bad but not good either. You were specifically told that equal volumes of equimolar solutions were mixed - what does it tell you about molar ratio of acid and base?

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  6. Mar 12, 2010 #5

    Char. Limit

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    Well... ammonium, I suppose.

    Wait, do the ammonium and hydroxide also form a precipitate? No, they can't... do they react to form ammonia and water themselves?

    I'm not sure if you are telling me whether I'm right or wrong, to be honest.



    Will do, right after this post.

    That they must be 1-to-1... Of course. Why I didn't see that before, I don't know.

    I'll get back to you on this one...

     
  7. Mar 12, 2010 #6

    Char. Limit

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    Ok, I've got another idea for c...

    Na2HPO4(aq) + HCl(aq) --> NaH2PO4(aq) + NaCl(aq)

    That's the only thing I can think of.

    And some confirmation for h...

    MnO4- is reduced to MnO2, right? And C2O42- is oxidized to CO2, right? But then where do the positive ions go... am I wrong here?
     
  8. Mar 12, 2010 #7

    Borek

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    Yes.

    Because to some extent whether you are right or wrong depends on what your teacher expects. Both approaches can be defended.

    And that's OK, although I would prefer it in the net ionic form.

    Depends on pH. As a rule of thumb you may remember that in basic solutions permanganate gets reduced to manganate (MnO4-), in neutral to MnO2, in acidic to Mn2+. Obviously you are not told anything about solution being basic nor acidified, so you can assume it is neutral.

    Yes.

    If they are not in the reaction, they are on the stand :wink:

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  9. Mar 12, 2010 #8

    Char. Limit

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    So, for c I have...

    H+(aq) + HPO42-(aq) --> H2PO4-(aq)

    And for h, which I'm still confused on...

    MnO4- + C2O42- --> MnO2 + 2 CO2 + 2 O2-

    The problem is, I'm almost certain that isn't right.
     
  10. Mar 12, 2010 #9

    Borek

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    And you are right about being wrong. No such thing as O2- in solution. In some crystalic oxides yes, but not in water.

    However, you are closing. You are right that you have excess oxygen. In redox when we have to balance oxygen and/or hydrogen, we do it using H2O/OH-/H+.

    Think what may have happen if O2- reacts with water.

    O2- + H2O ->

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    methods
     
  11. Mar 12, 2010 #10

    Char. Limit

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    So... if I add 2 H2O to both sides, I'd get four hydroxide ions in solution, right? But then it still isn't right: I have a 3- charge on the one side, and a 4- charge on the other. Can I just add an OH- ion to one side?
     
  12. Mar 13, 2010 #11

    Borek

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    No, you have to start finding correct ratio between permanganate and oxalate.

    At this moment you should be able to write correct skeleton equation, try to apply information from the pages I have linked to now. And if i can give you an advice - try the half reaction method. It may look less intuitive and more difficult, but is in general much more flexible and much closer to the reality.

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  13. Mar 13, 2010 #12
    a)--the carbon dioxide can't be aqueous, since no water is present.
     
  14. Mar 13, 2010 #13

    Borek

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    Good point, I skimmed too fast.

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  15. Mar 16, 2010 #14

    Char. Limit

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    I have one more question here.

    On ChemBuddy, half-reactions are said to be balanced by H in acidic solution and OH in basic solution. What do I use in neutral solution? Either one?
     
  16. Mar 16, 2010 #15

    Borek

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    Whichever suits you. And don't treat the idea too rigorously, sometimes it is much easier to use OH- in an acidic solution and H+ in basic. You can always use some combination of water/H+/OH- to modify the reaction so that correct ions are on the correct side.

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    methods
     
  17. Mar 16, 2010 #16

    Char. Limit

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    All right, I have what I think would be the correct reaction...

    2 MnO4-(aq) + 3 C2O42-(aq) + 8 H+(aq) --> 2 MnO2(s) + 4 H2O(l) + 6 CO2(g)
     
  18. Mar 17, 2010 #17

    Borek

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    Looks OK.

    Just remember that MnO2 being a product was predicted by a simple rule of thumb, so chemical reality can be different.

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