For the answer to the question above, first divide the
BaSo4 and 233.4 and it will look like this.
(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt
Then,
(10.0 g) / (0.0480 mol) = 208.3 g/mol
So the answer would be BaCl2.
I hope my answer helped you.
