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!