A10.00 gram sample of a soluble barium salt is treated with an excess of sodium sulfate to precipitate 11.21 grams baso4, (m= 233.4). which barium salt is it?

For the answer to the question above, first divide the 
BaSo4 and 233.4 and it will look like this.

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt 
Then,
(10.0 g) / (0.0480 mol) = 208.3 g/mol 
So the answer would be BaCl2.
I hope my answer helped you.