Đáp án:
Giải thích các bước giải:
`n_{Ba(OH)_2}=0,25.1=0,25(mol)`
`n_{Al_2(SO_4)_3}=(25,65)/342=0,075(mol)`
`3Ba(OH)_2+Al_2(SO_4)_3->2Al(OH)_3↓+3BaSO_4↓`
`(0,25)/3<(0,075)/1=>Ba(OH)_2` dư
`n_{Ba(OH)_2}=0,25-0,075.3=0,025(mol)`
`n_{Al(OH)_3}=2n_{Al_2(SO_4)_3}=0,15(mol)``
`2Al(OH)_3+Ba(OH)_2->Ba(AlO_2)_2+4H_2O`
`(0,15)/2>(0,025)/1=>Al(OH)_3` dư
`n_{Al(OH)_3dư}=0,15-0,025.2=0,1(mol)`
`n_{BaSO_4}=3n_{Al_2(SO_4)_3}=0,225(mol)`
`m=m_{Al(OH)_3dư}+m_{BaSO_4}=0,1.78+0,225.233=60,225(g)`