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Topic: Theoretical yield  (Read 10504 times)

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Offline mrorganiclover

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Theoretical yield
« on: December 16, 2014, 07:58:04 AM »
Good day all


Will someone please help me calculate the theoretical yield for the preparation of K2[Cu(C2O4)2].4H2O.

Dissolve 3.0g 2H2C2O4.2H2O in 40 cm^3 H2O.

Add 2.2g K2CO3.

Add Cu(II)O.

These are the reagents involved in the synthesis (Obviously I have left out the synthesis details).

I know that I have to write a balanced equation, which is the part giving me trouble.

So far I have:

2H2C2O4.2H2O + K2CO3 = H2CO3 + 2KHC2O4 + 4H2O.

This is where I have trouble. I have decided the K2CO3 is acting as a catalyst and the above equation is not required, but which equation would therefore be used?

All help greatly appreciated.






Offline Hunter2

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Re: Theoretical yield
« Reply #1 on: December 16, 2014, 08:08:26 AM »
You need two equations. One is to neutralize the copper oxide. The second one is to convert the Carbonate to Oxalate. Your approach was already on the right path, but develop for neutral potassium oxalate.

Offline mrorganiclover

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Re: Theoretical yield
« Reply #2 on: December 16, 2014, 08:29:32 AM »
Can you please elaborate a little.

Thanks

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Re: Theoretical yield
« Reply #3 on: December 16, 2014, 08:47:14 AM »
Try to predict products for these two reactions:

CuO + H2C2O4 :rarrow:

and

CuO + 2H2C2O4 :rarrow:

(note: don't touch stoichiometric coefficients on the left side, with correct products reaction will be already balanced).

Second reaction is close to being net ionic reaction for the whole process.
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Re: Theoretical yield
« Reply #4 on: December 16, 2014, 08:47:37 AM »
The second one is to convert the Carbonate to Oxalate.

Huh?
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Offline mrorganiclover

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Re: Theoretical yield
« Reply #5 on: December 16, 2014, 09:10:11 AM »
Thanks.

CuO + H2C2O4 :rarrow: CuC2O4 + 2H2O

CuO + 2H2C2O4 :rarrow: CuC2O4 + 2H2O + H2C2O4

What now?

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Re: Theoretical yield
« Reply #6 on: December 16, 2014, 09:58:45 AM »
CuO + H2C2O4 :rarrow: CuC2O4 + 2H2O

OK

Quote
CuO + 2H2C2O4 :rarrow: CuC2O4 + 2H2O + H2C2O4

Not exactly - you have an identical molecule of the oxalic acid on both sides, so it didn't react.

Take a look at the compound you should produce - it is an ionic salt. Can you split it into cations and an anion? Can you make this anion fit the product side of the reaction above?
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Offline Hunter2

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Re: Theoretical yield
« Reply #7 on: December 16, 2014, 10:04:09 AM »
The second one is to convert the Carbonate to Oxalate.

Huh?

Why HUh?

 We want to have this K2[Cu(C2O4)2]*4H2O.

K2CO3 + H2C2O4 => K2C2O4 + H2O + CO2


Together with the already developed CuC2O4  it can be combined to the product.

K2C2O4 + CuC2O4 +  4 H2O => K2[Cu(C2O4)2]*4H2O
« Last Edit: December 16, 2014, 10:16:46 AM by Hunter2 »

Offline mrorganiclover

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Re: Theoretical yield
« Reply #8 on: December 16, 2014, 10:13:17 AM »
Okay I think I may have it:

Cu^2+O^2- + H2C2O4^2-  :rarrow: (Cu^2+(C2O4^2-)2^2-

Offline Hunter2

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Re: Theoretical yield
« Reply #9 on: December 16, 2014, 10:18:57 AM »
No look above.

CuO + H2C2O4 =>  CuC2O4 + H2O

K2CO3 + H2C2O4 => K2C2O4 + H2O + CO2

Both combine

Offline mrorganiclover

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Re: Theoretical yield
« Reply #10 on: December 16, 2014, 11:07:33 AM »
Okay now having worked out the moles of each:

CuO = 0.01006 mol
H2C2O4 = 0.03503 mol
K2CO3 = 0.01447 mol


And deciding that the CuO is the limiting reagent.

What do I do now?

Offline Hunter2

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Re: Theoretical yield
« Reply #11 on: December 16, 2014, 11:16:38 AM »
Yes correct. So you can see that you need 0,01 mol CuO, the double amount of oxalic acid and the same amount of potassium carbonate to get the same mole of the product. So the carbonate fits, but your oxalic acid value is wrong. Check its the hydrate what is used!!!

Offline mrorganiclover

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Re: Theoretical yield
« Reply #12 on: December 16, 2014, 11:35:31 AM »
Fits where? Please excuse me, this is my first time calculating theoretical yield.

I have calculated the hydrate to be 0.02502 mol

Offline Hunter2

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Re: Theoretical yield
« Reply #13 on: December 16, 2014, 12:14:12 PM »
The given masses fit.
According the chemical equation you need 1 CuO : 2 Oxalic acid and 1  potassium carbonate. 1 :2 :1.

So you get 0.01 mol to 0.025 : 0,014

Offline mrorganiclover

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Re: Theoretical yield
« Reply #14 on: December 16, 2014, 12:30:21 PM »
Yes I understand this, but how is this information used to calculate theoretical yield?

Would I just plug the values in moles into the ratio?

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