BaCl2 reacts with KOH forming KCl which is a salt and Ba(OH)2 which is the precipitate.
The initial UNBALANCED equation expressing this reaction would be:
BaCl2 + KOH .............> KCl + Ba(OH)2
Now, we need to balance this equation:
we have two moles of Cl as reactants and one mole only as product. Therefore, we will multiply the KCl in the product by 2 and the KOH in the reactants by 2.
This will balance the equation as follows:
BaCl2 + 2KOH ......> 2KCl + Ba(OH)2
Noticing this equation, we will find that:
The precipitate was formed due to the combination of the Ba2+ ion with 2 OH- ions as follows:
Ba2+ + OH- ............> Ba(OH)2
