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Old 07-21-2008, 09:36 PM   #1
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eccchhhhhhh!!! Is this AP Chem? Dx

Here's what I did.

Originally Posted by Maki View Post

A 500mL bottle of concentrated sulfuric acid (18molL-1) was dropped in a laboratory accident. Solid sodium hydrogen carbonate (NaHCO3) was used to neutralize the spilled acid.

a) Justify the choice of the solid sodium hydrogen carbonate to clean up the spill. Include relevant equations(s).

H2SO4+NaHCO3=CO2+H2O+NaSO4

500ml of H2SO4 (sulfuric acid) with 18molL-1 is 9 mols.

Balancing the rest of the equation gives you:

9(H2SO4)+18(NaHCO3)=18(CO2)+9(H2O)+9(Na2SO4)

Your products are then CO2 (harmless), H2O (even more harmless), and Na2SO4 (a precipitate).



b) Calculate the minimum mass of sodium hydrogen carbonate needed to neutralise the spilled acid completely.




:3?
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Old 07-21-2008, 09:38 PM   #2
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Oh balancing equations? I hated that...
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Old 07-21-2008, 09:38 PM   #3
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balancing equations isnt hard
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Old 07-21-2008, 09:40 PM   #4
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Originally Posted by PoisoN View Post
balancing equations isnt hard
For me it was...

I used something called the EBA or something... I hate trial and error...
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Old 07-22-2008, 02:51 PM   #5
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Originally Posted by Ralath View Post
eccchhhhhhh!!! Is this AP Chem? Dx

Here's what I did.



:3?
Brings me back to high school chemistry.

I think Ralath got the products. I disagree slightly on the balanced equation:

H2SO4 (l) + 2NaHCO3 (s) -> Na2SO4 + 2H2O + 2CO2

(remember to reduce the coefficients to the simplest ratio)

Why can't we have a maths lesson? It's easier.
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Old 07-22-2008, 03:15 PM   #6
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Originally Posted by lamchopz View Post
H2SO4 (l) + 2NaHCO3 (s) -> Na2SO4 + 2H2O + 2CO2

(remember to reduce the coefficients to the simplest ratio)
Owned.
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Old 07-22-2008, 06:51 PM   #7
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Actually. I did see that.

But then I was having problems finding enough oxygens on both sides.

So... yeah. Dx

I was too excited to actually figure that out. Couldn't figure out what to do with the carbons for a long time...

So I just left it like it is.

But doesn't change part B.
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Old 07-23-2008, 02:26 AM   #8
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The oxidation number of each element hasn't changed so I'm guessing you used trials and errors.

What I did was first balance the sodium, then the hydrogen and finally the oxygen (consequently the carbon).

Part b is easy once you get the balanced equation.
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Old 07-23-2008, 03:20 AM   #9
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hrm. I can't remember what I did... lol..

Oh no wait. I know. I stuck in the 9 in front after I figured out the 9 moles. And then I just forgot to reduce. xD
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Old 07-23-2008, 03:32 AM   #10
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