An aqueous solution containing 1.728g of an impure mixture was analysed for barium ions by adding sufficient sulphuric 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 0.408g of dried barium sulphate was obtained?
I used this as the balanced equation but somehow it looks funny to me...
BaX + H2SO4 ------> BaSO4 + HX
The Attempt at a Solution
no of mols BaSO4 produced = mass/Mr
Mr of BaSo4= (137.3)+ (32.1) + (16x4)
BaX : BaSO4
1 : 1
mass Ba^2+ = molsx Mr
=(6.033x10^-4) x 137.3
% of Ba^2+ present in mixture= (0.0828/1.728) x100
Am i on the right track? Plz help!