[VkL]

Equal volumes of .200 M Ba(NO3)2 and .120 M K2Cr04 are mixed. If the precipitation of yellow Barium chromate is quantitative, what is the [Ba+2] in the final solution?
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This is what I got so far. I don't know what to do next:

Ba(NO3)2(aq) + K2(CrO4)(aq) <-----> Ba(CrO4) (s) + 2KN03(aq)
Ba +2 + NO3 - + K+ + CrO4 -2 <---> Ba(Cro4)(s) + 2 k + + 2 NO3 -     


There is NO [Ba+2] in the final solution
What do I do next?

Thank you for any help at all.

[Dan]

http://en.wikipedia.org/wiki/Limiting_reagent

[Arkcon]

OK, so you have equal volumes of solutions, at different concentrations, and you know they ionize.  But, how much of each reactant do you have?  How much gets used up?  And what's left over?

[VkL]

But How do I determine how many moles I have initially?
I don't have the volume?

[Arkcon]

They said equal volumes, so ... pick any one.  Say, a liter.  Then calculate.  The double it, and calculate again.  Then 10 times it.  Then you'll learn the relationship.  Doubling and 10x are easy multipliers to apply to any math problem.  And you learn what the proportional relationship is.

[VkL]

Thank you very much for your help.

SO I picked a 1 L total basis.

Therefore, I have 0.2 n/L * ( 0.5 L) = .1 mol Ba(No3)2
                         0.1 n/L * (0.5 L) =  .05 mol K2(CrO4)

Therefore, K2(CrO4) is the limiting reagent, since it is 1:1 with the other reactant (Ba(No3)2) and has lower moles.

Now do I find out how much insoluble product I make ? Ba(Cro4)(s)

Since reagents and product are 1:1:1,
I also have 0.05 mol of Ba(Cro4)(s),
which means I also would have 0.05 mol of Ba+2 that reacted.

Am I doing this right?

[Arkcon]

So, using the chemical punctuation for concentration, [], if [0.200 M] Ba(NO₃)₂ and [0.120 M]  K₂Cr0₄ (notice our forum has subscripts) react in equilmolar amounts according to the reaction, you can now give a [Ba²⁺] number for what's left over.

[VkL]

Thank you again, for your help.

I cannot see the problem to completion. I don't know why.

my basis is 20 L total ( 10 L for each reagent)

So,

[0.2] Ba(NO₃)₂ * 10 L = 2 mol Ba(NO₃)₂
[0.120] K₂Cr0₄ * 10 L = 1.2 mol K₂Cr0₄ (Limiting)

Since both reagents and Insoluble product are 1:1:1 ratio

I am stuck here. I don't know how to proceed.
It would be very appreciated if I can see a detailed answer,
So i can see it being done.

[VkL]

Actually, I think I got it.

I got .04 Mol Ba(NO3)2, since it's 1:1 with Ba+2, it should be the same.

I will confirm if this is correct tomorrow.

Thank you so much for your help.