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

According to the law of conservation of mass, the amount of BARIUM present of the reactants is the same as the amount present in the products (the precipitate). 

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt 

(10.0 g) / (0.0480 mol) = 208.3 g/mol 

So it must have been BaCl2, because the molar mass of Barium is 137 which leave 71 grams left. Since Barium is a +2 charge, it means the atom next to it must be twice. Chlorine mass is 35, which twice is 71

Less stable  hope this