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November 08, 2025, 04:07:37 am

Author Topic: Reaction?  (Read 1869 times)  Share 

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m@tty

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Reaction?
« on: February 20, 2010, 04:22:23 pm »
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What products come from the reaction of: , and ?

Thanks
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96.85

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Edmund

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Re: Reaction?
« Reply #1 on: February 20, 2010, 05:00:39 pm »
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Am I right?  :-\
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m@tty

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Re: Reaction?
« Reply #2 on: February 20, 2010, 05:12:09 pm »
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I don't know.

The only useful information was that was added to precipitate sulfate. ie. is a product.
I couldn't figure out what was going to happen with aluminium, that's why I asked.

EDIT:
The answers didn't give the equation, this is for gravimetric analysis, so they just gave the mass of the precipitate.

The full question was:
0.500 g of dissolved in water
0.500 g of dissolved in water
excess added

What is the mass of the precipitate, . The answer is 1.85g
« Last Edit: February 20, 2010, 09:26:36 pm by m@tty »
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Edmund

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Re: Reaction?
« Reply #3 on: February 20, 2010, 05:36:50 pm »
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Hmm... Aluminium chloride is insoluble
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chem-nerd

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Re: Reaction?
« Reply #4 on: February 20, 2010, 05:42:06 pm »
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n(BaSO4) = n(SO42-) = n(Na2SO4) + 3 x n(Al2(SO4)3)

It's the BaSO4 that will form a precipitate and the rest of the reactants will remain in solution.

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Re: Reaction?
« Reply #5 on: February 20, 2010, 06:03:45 pm »
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you would have 2 equations, one is sodium sulfate + barium chloride, and the other is aluminium sulfate + barium chloride.  Both reactions would give you a barium sulfate precipitate.  And then you would needa find the mass of each of the barium sulfates in both the reactions and add them together, that should give you 1.85 g! :)
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Studyinghard

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Re: Reaction?
« Reply #6 on: February 20, 2010, 06:34:24 pm »
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An easy method is ...

a) get mole of Na2SO4 and so that is the mole of SO42-

b) get mole of (Al2(SO4)3) and multiply that times 3 because there are 3 sulfate ions to get the mole of sulfate in this compound.

then you add the number of mole in a) and b) and that is the mole of SO42-

That is then the mole of BaSO4 because of the 1:1 to ratio.

then with the mole you figure out the mass of BaSO4
« Last Edit: February 20, 2010, 06:38:13 pm by Studyinghard »
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chem-nerd

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Re: Reaction?
« Reply #7 on: February 20, 2010, 06:38:58 pm »
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^^ that's exactly what I did here :)

n(BaSO4) = n(SO42-) = n(Na2SO4) + 3 x n(Al2(SO4)3)

It's the BaSO4 that will form a precipitate and the rest of the reactants will remain in solution.

Studyinghard

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Re: Reaction?
« Reply #8 on: February 20, 2010, 06:40:31 pm »
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^^ that's exactly what I did here :)

n(BaSO4) = n(SO42-) = n(Na2SO4) + 3 x n(Al2(SO4)3)

It's the BaSO4 that will form a precipitate and the rest of the reactants will remain in solution.
]
oh my bad ... as soon as i saw an equation i didnt read the post :P
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simonhu81292

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Re: Reaction?
« Reply #9 on: February 20, 2010, 07:22:14 pm »
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Hmm... Aluminium chloride is insoluble
hey edmund... isn't it only silver chloride and lead chloride aren't soluble ...
so aluminium chloride is also insoluble ... ?  :o
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Edmund

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Re: Reaction?
« Reply #10 on: February 20, 2010, 07:28:14 pm »
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Hmm... Aluminium chloride is insoluble
hey edmund... isn't it only silver chloride and lead chloride aren't soluble ...
so aluminium chloride is also insoluble ... ?  :o
Oh yeah I looked it up wrongly, Ag and not Al :P
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simonhu81292

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Re: Reaction?
« Reply #11 on: February 20, 2010, 07:34:44 pm »
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lol...
yeah ..matty ..
the answer is right if you follow chem-nerd's way !! ...just checked

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