Kết tủa lớn nhất nên $Al(OH)_3$ không tan.
$FeSO_4+Ba(OH)_2\to BaSO_4\downarrow + Fe(OH)_2\downarrow$
$\Rightarrow n_{BaSO_4}=n_{Fe(OH)_2}=0,040125(mol)$
$Al_2(SO_4)_3+3Ba(OH)_2\to 2Al(OH)_3\downarrow + 3BaSO_4\downarrow$
$n_{Al(OH)_3}=\dfrac{267}{34000}.2=\dfrac{267}{17000}(mol)$
$n_{BaSO_4}=3.\dfrac{267}{34000}=\dfrac{801}{34000}(mol)$
$\to m=0,040125.233+0,040125.90+\dfrac{267}{17000}.78+\dfrac{801}{34000}.233$
$=19,675g$