In this chemical reaction 15.5 g of KOH is given and 20.5 g of BaCl2 is given. How much precipitate is produced? How much excess is left after the reaction is complete?

Question

anonymous · Answer

$$2KOH + BaCl _{2}\rightarrow 2KCl + Ba(OH)_{2}$$

anonymous · Answer

Start by finding the number of moles of each reactant.

Then determine which is the limiting reactant...

anonymous · Answer

What did you get? For KOH i got 0.276 and for BaCl2 i got 0.0984

anonymous · Answer

That is correct. Which is the limiting reactant?

anonymous · Answer

BaCl2?

anonymous · Answer

Yes. How many moles of the precipitate would you expect to get?

anonymous · Answer

Sorry.  This website keeps dropping the connection.

anonymous · Answer

Its okay.