Using molarity in stoichiometry problems

AI Thread Summary
The discussion revolves around calculating the molarity of aluminum sulfite in a reaction with barium chloride. The user starts by balancing the chemical equation and recognizes the need to convert the volume of aluminum sulfite from liters to moles but lacks its molarity. There is confusion regarding the missing volume information for the barium chloride solution, which is necessary for completing the stoichiometric calculations. The user expresses uncertainty about how to proceed without additional data, suggesting that the problem may have been incorrectly stated by the teacher. Overall, the thread highlights the importance of having all relevant variables for stoichiometry problems.
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Homework Statement



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.

Homework Equations


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?
 
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That's all the variables you have? It doesn't say anything about the volume of Barium Chloride or any of the products? I may be missing something as I just woke up, but I can't see what to do without something else.
 
yeah that's all it gave me...
 
I would just assume that I had 3.00 mol Barium Chloride then. There has to be another variable...
 
lol, the teacher had the problem wrote wrong
 
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