# Limiting Reagent Stoichiometry

What mass of precipitate forms when a solution containing 6.24 g of potassium sulfide is reacted with a solution containing 19.2 g of barium nitrate?

I have already identified the limiting reagent $\left(\text{K}_2 \text{S}\right)$ as well as the mass of the precipitate.

My question, however, is: why is the Barium Sulfide formed in the product a solid and not aqueous? I thought the problem translated into the molecular equation: $$\text{K}_2 \text{S}_{\text{(aq)}} + {\text{Ba}(\text{NO}_3)_2}_{\text{(aq)}} \ce {->} 2{\text{KNO}_3}_{\text{(aq)}} + \text{BaS}_{\text{(aq)}}$$

But according to the answer key I was given, $\text{BaS}$ is a solid precipitate, not aqueous. How can that compound be a solid if, according to solubility rules, all sulfides plus an alkali earth metal are soluble? Shouldn't it be aqueous, not solid?

• You're right, it is soluble, given enough water. However, potassium nitrate is WAYYY more soluble, so it will dissociate first. If there's not enough water for both of them to dissociate, the barium sulfide will precipitate out first. Feb 5, 2015 at 14:54
• Oh, okay. So in general, when doing these types of questions, what guidelines should I abide by to figure out if a compound in the product is aqueous or not (other than solubility rules)? Feb 5, 2015 at 15:10
• Did the question specify which product was the precipitate? Feb 5, 2015 at 18:20
• @wes3449 it didn't, unfortunately. the question was what i typed verbatim Feb 5, 2015 at 19:32
• See my answer for a moderately detailed explanation. Feb 5, 2015 at 19:52

$\ce{2BaS2 + 2H2O -> Ba(OH)2 + Ba(SH)2}$