Question

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?

Answer 1

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

Answer 2

Start by finding the number of moles of each reactant.

Then determine which is the limiting reactant...

Answer 3

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

Answer 4

That is correct. Which is the limiting reactant?

Answer 5

BaCl2?

Answer 6

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

Answer 7

Sorry. This website keeps dropping the connection.

Answer 8

Its okay.