[Jabby64]

An aqueous solution containing 1.728 g of an impure mixture was analyzed for barium ions by adding sufficient sulfuric acid solution to completely precipitate the barium ions as its sulphate. The precipitate was filtered, washed and dried to constant mass and finally weighed. What is the mass percent of barium ions in the mixture, if 0.1408 g of dried barium sulphate was obtained?

[sjb]

An aqueous solution containing 1.728 g of an impure mixture was analyzed for barium ions by adding sufficient sulfuric acid solution to completely precipitate the barium ions as its sulphate. The precipitate was filtered, washed and dried to constant mass and finally weighed. What is the mass percent of barium ions in the mixture, if 0.1408 g of dried barium sulphate was obtained?

How many moles of barium sulphate were precipitated?

[Jabby64]

An aqueous solution containing 1.728 g of an impure mixture was analyzed for barium ions by adding sufficient sulfuric acid solution to completely precipitate the barium ions as its sulphate. The precipitate was filtered, washed and dried to constant mass and finally weighed. What is the mass percent of barium ions in the mixture, if 0.1408 g of dried barium sulphate was obtained?

I don’t know if you mean this:
RMM Ba2SO4 =(2*137)+(1*32)+(4*16)=370
Moles= Mass/RMM
            =0.1408/370
            =3.8*10-4mol Ba2SO4

[Borek]

Ba2SO4

Check the formula.