Question

A 10.00 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?

Answer 1

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