Precipitate mass question!

  • Thread starter melxo
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  • #1
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I have a question in my Chemistry homework that has to do with precipitates. Theres nothing in my notes what so ever and I ont remember my prof going over any of this. Any help would be appreciated, thanks!

Question:

24.8 g of BA(No3)2 were dissolved in enough water to make 150mL of solution. To this, 25 mL of 0.329 M Na2So4 was added and a precipitate formed.

What is the maximum amount of precipitate, in grams, that can form in this reaction?






so far I have...

moles Ba(NO3)2=24.8 g (1 mole/261.32g)=.0949

moles NA2(SO4)= .025L(.329M)=.008225 mol which is the limiting reactant

now what do i do?! and please dont just give me the answer!
 

Answers and Replies

  • #2
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What is the precipitate formed?
What is the ratio of reactants to products? (e.g. if A + 2B --> B2A, then B : B2A is 2 : 1)
Knowing this you can solve.
 
  • #3
Borek
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Start with the reaction equation, this is simple stoichiometry.

And post this type of the questions in the homework subforum. I am moving the thread.
 

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