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Topic: Precipitate and Net Ionic Equations  (Read 4308 times)

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scfan000

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Precipitate and Net Ionic Equations
« on: June 04, 2006, 07:12:12 PM »
So I have this problem on a packet and
have a slight idea to how to solve it. but am completely confused on how to ???

A chemist has a 3.00 g mixture of two solids, sodium sulfate and potassium chloride.  They are placed in enough water to dissolve completely.  Then enough barium nitrate solution is added to precipate out all applicable ions.  THis precipate was filtuered out of solution.  The preciptate had a mass of 2.14 g.

What was the original percent by mass of each sodium sulfate and potassium chloride in the mixture?

Sodium Sulfate: Na2SO4
Potassium Chloride: KCl
Barium Nitrate: BaNO3

I have absolutely no idea how to solve the problem because I was confused to a model answer given from a similar problem
which assumed one of the elements (in solving) as being 3.00 g to find the preciptate.

Plz Help :-\

Offline Yggdrasil

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Re: Precipitate and Net Ionic Equations
« Reply #1 on: June 04, 2006, 08:15:03 PM »
First start out by writing a balanced chemical equation.

Offline anarchron

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Re: Precipitate and Net Ionic Equations
« Reply #2 on: June 04, 2006, 08:34:01 PM »
Firstly you need to determine the formula of the percipitate.

As BaSO4 is not very soluble, that would be the percipitate. Once you know that, ignore the KCl part. The formula should look something like:

Na2SO4(aq) + Ba(NO3)2 -> BaSO4(s) + 2NaNO3(aq)

once that's done, it's a simple matter of

n(BaSO4) = 2.14/233.3 = 9.172x10-3mol

It is a 1:1 reaction, therefore:

n(Na2SO4) =  9.172x10-3mol

mass(Na2SO4) =  9.172x10-3mol *142
= 1.30g

hence:

mass(KCl) = 3.00 -1.30 = 1.70g

Voila!

scfan000

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Re: Precipitate and Net Ionic Equations
« Reply #3 on: June 04, 2006, 08:39:07 PM »
ty ty  :D

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