$4A+nO_2\xrightarrow{t^o}2A_2O_n$
`4A` `4A+32n`
`4,05` `7,65`
Theo PT
`=>7,65.4A=4,05(4A+32n)`
`=>30,6A=16,2A+129,6n`
`=>A=9n`
`n=1=>A=9` (loại)
`n=2=>A=18` (loại)
`n=3=>A=27 Al`
`b,`
`n_(Al)=\frac{4,05}{27}=0,15(mol)`
$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$
`0,15` `0,075`
`Al_2O_3+6HCl->2AlCl_3+3H_2O`
`0,075` `0,45`
`m_(HCl)=0,45.36,5=16,425(g)`
`m_(dd HCl)=\frac{16,425}{20%}=82,125(g)`