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Topic: Writing the chemical formula for water in a reaction with an aqueous solution  (Read 2705 times)

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Offline matermultorum

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I've noticed that most of the time in a chemical reaction equation with an aqueous solution, you don't write the H2O that is in the solution in the equation.

If an electric current is passed through aqueous solution of sodium chloride Chloride and hydrogen are produced at different electrodes. If the liquid is then evaporated from the mixture, a residue of sodium hydroxide remains. Write a balanced chemical equation.

I have the correct answer. NACL + H2:rarrow: NA(OH)2 + H2+Cl2

Is there a general rule about when to write the H2O molecule in the equation?  IS it only when it is mentioned as a separate reactant or product?

Thanks

Offline Arkcon

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Yes.  That's correct.  The electrolysis drives a red-ox reaction.  And water is reacted, along with NaCl.

Here's one for you:  What if there is no water?  Consider the electrolysis of molten sodium chloride NaCl (l).  What happens?
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Borek

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I have the correct answer. NACL + H2:rarrow: NA(OH)2 + H2+Cl2

Note: it is not a correct equation. Apart from the fact you ignored the capitalization formula of sodium hydroxide is different from what you wrote. Plus, it is not balanced.
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Offline matermultorum

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I have the correct answer. NACL + H2:rarrow: NA(OH)2 + H2+Cl2

Note: it is not a correct equation. Apart from the fact you ignored the capitalization formula of sodium hydroxide is different from what you wrote. Plus, it is not balanced.

2NaCl+ 2H2:rarrow: 2Na(OH)2 + H2+Cl2

Sorry, I should have previewed that better.  I think it is balanced now and the typo has been fixed.  If the Sodium Chloride is in molten form and electrolysis is performed, then would you end up with sodium and chloride ions? Or would there be no reaction?

Here are 3 equations all involving aqueous solutions (taken from the text 5th ed. Introductory Chemistry by Zumdahl)

In the first one

"Solid potassium reacts with liquid water to form gaseous hydrogen and potassium hydroxide that dissolves in the water" 

So I think I understand here that the water in the aqueous solution is included on the right not as H2O But in the designation (aq). H2O is listed as a reactant so it is listed on the left.

2K(s)+2H2O (l)  :rarrow: H2 (g) + 2KOH (aq)

solid zinc is added to an aqueous solution of hydrogen chloride (which is called hydrochloric acid). The products are gaseous hydrogen and aqueous zinc chloride. 

So since there is an aqueous solution on both sides the water is included in the designation (aq) and isn't part of the reaction

Zn(s) + 2HCl(aq)  :rarrow: H2 (g) + ZnCl2 (aq)

Hydrofluoric acid reacts with silicone in glass to form gaseous silicone tetrafluoride and liquid water

Water is listed as a product so it is included on the right. It is part of the (aq) designation

SiO2 (s) + 4HF (aq)  :rarrow: SiF4 (g) + 2H2O (l)

In the one I posted originally, water isn't mentioned as a reactant, but is included on the reactant side. It is mentioned as something that evaporates after the reaction. So sometimes when something is listed as aq you do put a separate water molecule in the equation and sometimes you don't. Maybe at this point we should just take our best guess because we don't have enough knowledge to do anything else?

One thing I think I am noticing as I teach through this book with two of my daughters, is that chemistry isn't quite so sequential as math is and that in order to teach concept A when concept B is involved but you can kind of get by(and need to) without explaining concept B. If you tried to teach Concept B first you would run into the same issue.  You can't thoroughly explain every part that might be involved in one chapter.

We have not leaned about red-ox or any of the other aqueous reactions. We are in the chapter about balancing chemical equations and the previous chapter was nomenclature.

Thanks for the help




Offline Borek

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I think it is balanced now

First - formula for sodium hydroxide is still wrong. Second - have you counted hydrogen atoms on both sides of the equation?

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If the Sodium Chloride is in molten form and electrolysis is performed, then would you end up with sodium and chloride ions?

NaCl is made of sodium and chloride ions, that's not something you can end with if there is a reaction (and there will be one).

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So I think I understand here that the water in the aqueous solution is included on the right not as H2O But in the designation (aq). H2O is listed as a reactant so it is listed on the left.

2K(s)+2H2O (l)  :rarrow: H2 (g) + 2KOH (aq)

I would say it is more like water that is used to dissolve produced KOH is ignored on the left, but the equation is OK.

(two other equations omitted as they are OK).

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In the one I posted originally, water isn't mentioned as a reactant

You can't produce hydrogen form just NaCl, so it is obvious that water must be a reactant, no need to take guesses.

Quote
One thing I think I am noticing as I teach through this book with two of my daughters, is that chemistry isn't quite so sequential as math is and that in order to teach concept A when concept B is involved but you can kind of get by(and need to) without explaining concept B. If you tried to teach Concept B first you would run into the same issue.  You can't thoroughly explain every part that might be involved in one chapter.

We have not leaned about red-ox or any of the other aqueous reactions. We are in the chapter about balancing chemical equations and the previous chapter was nomenclature.

I feel your pain. Math is a theoretical construct which builds on facts that have to be established earlier, chemistry is a real life thing, and as such is sometimes difficult to systematize, often you need several separate ideas to explain what is going on.

But you don't need to know what redox is to balance the reaction equation. Redox reactions are just subset of reactions, specific one, but basically governed by the same rules every other reaction is. And as I pointed out above, whether we include water in the reaction equation basically depends on whether the water itself reacts (is produced) or not.

Yes, it can get tricky, but not in the cases you have listed so far.
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