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Offline jsmith613

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Displacement equations
« on: September 06, 2009, 08:46:03 AM »
I understand how to do displacement equations where the molecules reacting have 2 ions
e.g: K2CO3 + NaOH --> KOH + Na2CO3

However I do not know how to do displacement equations where one of the molecules reacting has 3 ions and the other one has 2
e.g: C6H8O7 + K2CO3 --> ??

OR where both molecules reacting have 3 ions
e.g: K2Cr2O7 + C6H8O7 --> ??

(I am not sure it equations 2 and 3 are valid but i hope you get the idea)

Please can someone explain to me how to do equations like 2 and 3

Offline cliverlong

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Re: Displacement equations
« Reply #1 on: September 06, 2009, 09:10:02 AM »
I understand how to do displacement equations where the molecules reacting have 2 ions
e.g: K2CO3 + NaOH --> KOH + Na2CO3
Do group 1 carbonates react with alkalis? I haven't come across this before.

Please use the subscript button on the editor. It makes the formulae much more readable e.g. K2CO3 becomes K2CO3 with little effort.
Quote
However I do not know how to do displacement equations where one of the molecules reacting has 3 ions and the other one has 2
e.g: C6H8O7 + K2CO3 --> ??

OR where both molecules reacting have 3 ions
e.g: K2Cr2O7 + C6H8O7 --> ??

(I am not sure it equations 2 and 3 are valid but i hope you get the idea)

Please can someone explain to me how to do equations like 2 and 3

I do not recognise C6H8O7 (or C6H8O7). Being comprised of just carbon, hydrogen and oxygen the compound is going to be covalent not ionic.

First you need to find out the molecular formula for this compound. For example: methylamine: CH3NH2 gives some information about the bonding between the atoms whereas CNH5 is pretty hopeless. The more atoms you get the more useless a simple listing of atoms with a single subscript becomes.

You need a more informative formula than : C6H8O7. For the number of carbons it has a relatively high number of oxygens which need to bond to the carbons as either alcohol or carboxyl groups. Perhaps it is some kind of sugar? http://www.rpi.edu/dept/chem-eng/Biotech-Environ/FUNDAMNT/rings.htm
However, your formula is a bit "short" on hydrogens. I just don't recognise it - so it is difficult to determine if it has functional groups that will react with K2CO3 .

Clive

Offline Borek

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Re: Displacement equations
« Reply #2 on: September 06, 2009, 09:19:52 AM »
No idea what you mean by "have 3 ions".
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Offline Arctic-Nation

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Re: Displacement equations
« Reply #3 on: September 06, 2009, 09:43:57 AM »
The weird oxygen compound is probably citric acid, and having 3 ions means tribasic. You not being able to figure out that last one disappoints me, Borek. ;)

In cases where the ratio between the numbers of ionizable positions isn't unity, you just need to balance it out like in any other reaction. In case of K2CO3 + NaOH gives Na2CO3 + KOH, the balanced equation is K2CO3 + 2 NaOH gives Na2CO3 +2 KOH (note that this reaction isn't very likely to happen, but it'll do as an example). When you have a ratio of 3:2 as in the second example, the balanced equation is 2A + 3B gives 2C + 3D.

Offline jsmith613

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Re: Displacement equations
« Reply #4 on: September 06, 2009, 10:58:57 AM »
Quote
I understand how to do displacement equations where the molecules reacting have 2 ions
e.g: K2CO3 + NaOH --> KOH + Na2CO3
Do group 1 carbonates react with alkalis? I haven't come across this before.
As I said I was not sure if the equation was correct it was just a very bad example to pick

Quote
However I do not know how to do displacement equations where one of the molecules reacting has 3 ions and the other one has 2
e.g: C6H8O7 + K2CO3 --> ??

OR where both molecules reacting have 3 ions
e.g: K2Cr2O7 + C6H8O7 --> ??

(I am not sure it equations 2 and 3 are valid but i hope you get the idea)

Please can someone explain to me how to do equations like 2 and 3

Quote
I do not recognise C6H8O7

First you need to find out the molecular formula for this compound.

C6H8O7 is citric acid
http://en.wikipedia.org/wiki/Citric_acid
------------------------
but the point is, is there a specific method to do reactions where there are 3 ions per molecule.

Offline sjb

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Re: Displacement equations
« Reply #5 on: September 06, 2009, 12:17:50 PM »
C6H8O7 is citric acid
http://en.wikipedia.org/wiki/Citric_acid
------------------------
but the point is, is there a specific method to do reactions where there are 3 ions per molecule.

Why could C6H8O7 not be Isocitric acid? http://en.wikipedia.org/w/index.php?title=Isocitric_acid&oldid=294467125. But that's not important, in this case.

What I think you may need to do is see what parts of the chemicals are changing, and which are not. In some cases all 3 ions may undergo redox, in others fewer.

Offline jsmith613

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Re: Displacement equations
« Reply #6 on: September 06, 2009, 01:34:19 PM »
Quote
What I think you may need to do is see what parts of the chemicals are changing, and which are not. In some cases all 3 ions may undergo redox, in others fewer.

In other words do you want find the equation and then analyze the products to work out how the reactions work 

Offline Borek

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Re: Displacement equations
« Reply #7 on: September 06, 2009, 02:50:03 PM »
C6H8O7 is citric acid

No, it is one of tens of compounds that have this formula. In the case of organic compounds it is structure that counts, just giving overall formula is ambiguous.

Quote
but the point is, is there a specific method to do reactions where there are 3 ions per molecule.

Still no idea what you mean by that. Citric acid dissociates to create four ions.
« Last Edit: September 06, 2009, 03:08:00 PM by Borek »
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Offline jsmith613

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Re: Displacement equations
« Reply #8 on: September 06, 2009, 04:01:33 PM »
Quote
but the point is, is there a specific method to do reactions where there are 3 ions per molecule.

Quote
Still no idea what you mean by that. Citric acid dissociates to create four ions.

What I mean is, how can one work out the products in a reaction such as: C6H8O7 + K2CO3 --> ??

Offline Borek

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Re: Displacement equations
« Reply #9 on: September 06, 2009, 05:42:03 PM »
There is no specific method. You have to know what the substances are and how they can react. In this case you have an acid reacting with carbonate.
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Offline jsmith613

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Re: Displacement equations
« Reply #10 on: September 07, 2009, 10:26:03 AM »
There is no specific method. You have to know what the substances are and how they can react. In this case you have an acid reacting with carbonate.

Yes but how do I know that name of the salt

Offline cliverlong

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Re: Displacement equations
« Reply #11 on: September 07, 2009, 10:26:06 AM »

C6H8O7 is citric acid
http://en.wikipedia.org/wiki/Citric_acid
------------------------
but the point is, is there a specific method to do reactions where there are 3 ions per molecule.
Aha!

Can anyone confirm if an unambiguous and correct way to write the (condensed) structural formula for this compound is

CH2COOHCOHCOOHCH2COOH    ?

So (as already commented) you have an acid and are reacting this with a carbonate. Do you know any reactions of this type? Can you write an equation for the reaction? Big hint: compare the equations of the reaction of hydrochloric acid and sulphuric acid with metal carbonates.

Note, citric acid can exist as a crystalline solid or can dissolve in water. Will either of these affect citric acids reactivity with a carbonate?

What ions can citric acid form, under what conditions, and what reactions does this mean it can participate in? Have you studied acid dissociation constants for monoprotic and polyprotic acids?

Clive

Offline jsmith613

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Re: Displacement equations
« Reply #12 on: September 07, 2009, 10:28:21 AM »
Quote

Aha!

Can anyone confirm if an unambiguous and correct way to write the (condensed) structural formula for this compound is

CH2COOHCOHCOOHCH2COOH    ?

So (as already commented) you have an acid and are reacting this with a carbonate. Do you know any reactions of this type? Can you write an equation for the reaction? Big hint: compare the equations of the reaction of hydrochloric acid and sulphuric acid with metal carbonates.

Note, citric acid can exist as a crystalline solid or can dissolve in water. Will either of these affect citric acids reactivity with a carbonate?

Clive

Acid + Carbonate --> Salt + water + CO2 (what is the name of the salt)

Quote
What ions can citric acid form, under what conditions, and what reactions does this mean it can participate in? Have you studied acid dissociation constants for monoprotic and polyprotic acids?
I have not studied

Offline Borek

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Re: Displacement equations
« Reply #13 on: September 07, 2009, 02:29:34 PM »
No magic here:

citric acid -> citrate
acetic acid -> acetate
formic acid -> formate

sulfuric acid -> sulfate
nitric acid -> nitrate
phosphoric acid -> phosphate
carbonic acid -> carbonate

Doesn't mean it is always that simple, salt of hydrochloric acid is chloride...
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Offline jsmith613

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Re: Displacement equations
« Reply #14 on: September 07, 2009, 04:36:39 PM »
No magic here:

citric acid -> citrate
acetic acid -> acetate
formic acid -> formate

sulfuric acid -> sulfate
nitric acid -> nitrate
phosphoric acid -> phosphate
carbonic acid -> carbonate

Doesn't mean it is always that simple, salt of hydrochloric acid is chloride...

OK so Potassium Carbonate + Citric acid --> Potassium Citrate + CO2 + Water
but thats not really what i care about, I am more interested as to how to determine the formula or Potassium Citrate
K2CO3 + C6H8O7 --> ??? + CO2 + H2O

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