`n_{Mg}=12/24=0,5(mol)`

`PTHHMg+2HCl->MgCl_2+H_2`

`(mol)` `0,5->`   `1`            `0,5`     `0,5`

`a.C_M HCl=1/{0,3}~~3,33(M)`

`b.V_{H_2}=0,5.24,79=12,395(l)`

`c.V_{dd sau}=300ml=0,3l`

`C_M MgCl_2 ={0,5}/{0,3}~~1,67(M)`

PTHH : `Mg + 2HCl \to MgCl_2 + H_2`

`n_{Mg} = \frac{12}{24} = 0,5` `mol`

`a)`

Theo phương trình : `n_{HCl} = 2n_{Mg} = 1` `mol`

`=> C_MHCl = \frac{1}{0,3} = 3,33` `M`

`b)`

Theo phương trình : `n_{H_2} = n_{Mg} = 0,5` `mol`

`=> V_{H_2} = 0,5 . 24,79 = 12,395` `l`

`c)`

Theo phương trình : `n_{MgCl_2} = n_{Mg} = 0,5` `mol`

`=> C_MMgCl_2 = \frac{0,5}{0,3} = 1,67` `M`

$#1212$