a) PTHH:
$\rm Mg + 2HCl \longrightarrow MgCl_2 + H_2 \uparrow$

b) $200\rm ml = 0,2(\rm l)$

$\Rightarrow n_{\rm HCl} = C_M . V = 1 . 0,2 = 0,2(\rm mol)$
Theo PTHH:

$n_{\rm H_2} = \dfrac{1}{2}n_{\rm HCl} = \dfrac{1}{2} . 0,2 = 0,1(\rm mol)$

$\Rightarrow V_{\rm H_2} = n_{\rm H_2}. 24,79 = 0,1 . 24,79 = 2,479(\rm l)$

c) Theo PTHH:
$n_{\rm Mg} = n_{\rm H_2} = 0,1(\rm mol)$

$\Rightarrow m_{\rm Mg} = n_{\rm Mg} . M_{\rm Mg} = 0,1 . 24 = 2,4(\rm g)$

d) $V_{\rm \text{dd X}} = V_{\rm \text{dd HCl}} = 0,2(\rm l)$

Chất có trong dung dịch $X$ là $\rm MgCl_2$
Theo PTHH:
$n_{\rm MgCl_2} = n_{\rm H_2} = 0,1(\rm mol)$

$\Rightarrow C_{M(\rm MgCl_2)} = \dfrac{n_{\rm MgCl_2}}{V_{\rm \text{dd X}}} = \dfrac{0,1}{0,2} = 0,5(\rm M)$

`\text{Ta có:}` `n_(HCl) = 0,2 * 1 = 0,2` `(mol)`

         `Mg + 2HCl -> MgCl_2 + H_2 uparrow`

`mol`  `0,1` $\leftarrow$  `0,2`  `->`  `0,1`  `->`  `0,1`  

`\color{#8C63A4}{\text{b)}}` `\text{Ta có:}` `V_(H_2) = 0,1 * 24,79 = 2,479` `(l)`

`\color{#8C63A4}{\text{c)}}` `\text{Ta có:}` `m_(Mg) = 0,1 * 24 = 2,4` `(g)`

`\color{#8C63A4}{\text{d)}}` `\text{Ta có:}` `C_(M_(MgCl_2)) = (0,1)/(0,2) = 0,5` `(M)`