A 0.670 g sample of impure Al2(SO4)3 reacts with excess BaCl2. If the sample produces 0.760 g of BaSO4, what is the mass percent of Al2(SO4)3 in the sample?

Answer 1

Al2(SO4)3 + 3BaCl2 ==> 3BaSO4 + 2AlCl3
How many mols BaSO4 were collected? That's 0.760/molar mass BaSO4 = approx 0.0033 but you need to do it more accurately.
Using the coefficients in the balanced equation convert mols BaSO4 to mols Al2(SO4)3.
Convert mols Al2(SO4)3 to grams. g \ mols x molar mass
Then % Al2(SO4)3 = [g Al2(SO4)3/mass sample)]8100 = ?