Theo gt ta có: [tex]n_{Al}=0,4(mol)[/tex]
PTHH: [tex]2Al+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2(1)\\H_2SO_4+Ba(OH)_2\rightarrow BaSO_4+2H_2O(2)\\Al_2(SO_4)_3+3Ba(OH)_2\rightarrow 2Al(OH)_3+3BaSO_4(3)\\2Al(OH)_3+Ba(OH)_2\rightarrow Ba(AlO_2)_2+4H_2O(4)[/tex]
Từ (1) suy ra dung dịch X bao gồm: [tex]0,2mol[/tex] $H_2SO_4$; $0,2mol$ $Al_2(SO_4)_3$
Từ (2); (3); (4) suy ra kết tủa bao gồm: $0,2mol$ $Al(OH)_3$; $0,8mol$ $BaSO_4$
[tex]\Rightarrow m=202(g)[/tex]