Giải thích các bước giải:

`Mg + 2HCl -> MgCl_2 + H_2 (1)`

`2Al + 6HCl -> 2AlCl_3 + 3H_2 (2)`

`n_{Mg}=a/24 (mol)`

`n_{Al}=a/27(mol)`

`TH1:` kim loại hết

theo `PTHH`

`n_{H_2 (1)} = n_{Mg}=a/24 (mol)`

`n_{H_2 (2)} = 3/2 .n_{Al}=a/18 (mol)`

`a/24  < a/18`

`-> n_{H_2 (1)} < n_{H_2 (2)}`

`TH2:` kim loại dư

`n_{H_2 (1)}=1/2 .n_{HCl}`

`n_{H_2 (2)}=1/2 .n_{HCl}`

`-> n_{H_2 (1)}= n_{H_2 (2)}`

`TH3:Mg` dư; `Al` hết 

`n_{H_2 (2)}= 3/2 .n_{Al}= a/18 (mol)`

`n_{HCl}=2.n_{H_2 (2)}=2 . a/18 = a/9 (mol)`

`-> `n_{H_2 (1)} = 1/2 .n_{HCl} = 1/2 .a/9 = a/18 (mol)`

`-> n_{H_2 (1)} = n_{H_2 (2)} = a/18`

`-` `n_{Al} = \frac{a}{27}` `(mol)`

`-` `n_{Mg} = \frac{a}{24}` `(mol)`

Phương trình hóa học: `Mg + 2HCl -> MgCl_2 + H_2\uparrow` `(1)`

`2Al + 6HCl -> 2AlCl_3 + 3H_2\uparrow` `(2)` 

+) Nếu kim loại hết: `n_{H_2(1)} = n_{Mg} = \frac{a}{24}` `(mol)`

`n_{H_2(2)} = \frac{a}{18}` `(mol)`

`->` `n_{H_2(1)} < n_{H_2(2)}`

+) Nếu axit hết: `n_{H_2(1)} = \frac{1}{2}.n_{HCl}`

`n_{H_2(2)} = \frac{1}{2}.n_{HCl}`

`->` `n_{H_2(1)} = n_{H_2(2)}`

+) Nếu `Mg` dư, `Al` hết: `n_{H_2(1)} = \frac{a}{18}` `(mol)`

`n_{H_2(2)} = \frac{1}{2}n_{HCl} = \frac{1}{2}. (\frac{a}{18}.2) = \frac{a}{18}` `(mol)`

`->` `n_{H_2(1)} = n_{H_2(2)}`