Đáp án:
Giải thích các bước giải:
`n_{Al(OH)_3}=(21,06)/78=0,27(mol)`
`H=80%`
`n_{Al(OH)_3pư}=0,27.80%=0,216(mol)`
$2Al(OH)_3\xrightarrow{t^o}Al_2O_3+3H_2O$
`n_{Al_2O_3}=0,5n_{Al(OH)_3}=0,108(mol)`
`m_{Al_2O_3}=0,108.102=11,016(g)`
`n_{H_2O}=1,5n_{Al(OH)_3pư}=0,324(mol)`
`m_{H_2O}=0,324.18=5,832(g)`