What is the molarity of 50.0mL sample of aluminum sulfite when it is reacted with 3.00M solution of barium chloride? The product of this double replacement is aluminum chloride and barium chloride.
The first thing i did was balance(all number after forumula is subscript):
Al2(SO3)3 + 3BaCl2 -> 2AlCl3 + 3BaSO3
molarity is :
Mol of solute/ L of solution
It gives me the volume so i need to find the mol
The Attempt at a Solution
as my teacher said, its best to always start with the value other than M. So:
.050L Al2(SO3)3 * something to take it out of liter and into mols. But in order to do that i need to know Al2(SO3)3's molarity, and i only know barium chlorides's.
I must be not considering something pertaining to barium chloride, but I'm not sure, can someone lend me a hand?