How Do You Write Chemical Equations for These Reactions?

[Char. Limit]

Homework Statement

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.

Homework Equations

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

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) + PO₄^3-(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) + C₂O₄^2-(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)

 

[Borek]

c. HPO₄^2- 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

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[Char. Limit]

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...

 

[Borek]

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

What other ions are in the solution?

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

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.

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?

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

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...

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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[Char. Limit]

What other ions are in the solution?

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?

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.

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

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

Will do, right after this post.

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?

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...

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buffer calculator, concentration calculator
pH calculator, stoichiometry calculator

 

[Char. Limit]

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 C₂O₄^2- is oxidized to CO₂, right? But then where do the positive ions go... am I wrong here?

 

[Borek]

do they react to form ammonia and water themselves?

Yes.

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

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

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

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

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

MnO₄^- is reduced to MnO₂, right?

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.

And C₂O₄^2- is oxidized to CO₂, right?

Yes.

But then where do the positive ions go... am I wrong here?

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

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[Char. Limit]

So, for c I have...

H⁺(aq) + HPO₄^2-(aq) --> H₂PO₄^-(aq)

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

MnO₄^- + C₂O₄^2- --> MnO₂ + 2 CO₂ + 2 O^2-

The problem is, I'm almost certain that isn't right.

 

[Borek]

And you are right about being wrong. No such thing as O^2- 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 O^2- reacts with water.

O^2- + H₂O ->

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methods

 

[Char. Limit]

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?

 

[Borek]

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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[PhaseShifter]

a)--the carbon dioxide can't be aqueous, since no water is present.

 

[Borek]

Good point, I skimmed too fast.

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[Char. Limit]

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?

 

[Borek]

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

 

[Char. Limit]

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

2 MnO₄^-(aq) + 3 C₂O₄^2-(aq) + 8 H⁺(aq) --> 2 MnO₂(s) + 4 H₂O(l) + 6 CO₂(g)

 

[Borek]

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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methods