Giải thích các bước giải:

Gọi a và b lần lượt là số mo, của Mg và MgO

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}(1)\\
MgO + 2HCl \to MgC{l_2} + {H_2}O(2)\\
NaOH + HCl \to NaCl + {H_2}O(*)\\
{m_{HCl}} = \dfrac{{400 \times 7,3\% }}{{100\% }} = 29,2g\\
 \to {n_{HCl}} = 0,8mol\\
{m_{NaOH}} = \dfrac{{100 \times 4\% }}{{100\% }} = 4g\\
 \to {n_{NaOH}} = 0,1mol\\
 \to {n_{HCl}}(*) = {n_{NaOH}} = 0,1mol\\
 \to {n_{HCl}}(1) + (2) = 0,7mol\\
\left\{ \begin{array}{l}
24a + 40b = 11,6\\
2a + 2b = 0,7
\end{array} \right. \to \left\{ \begin{array}{l}
a = 0,15\\
b = 0,2
\end{array} \right.\\
 \to {n_{Mg}} = 0,15mol\\
 \to {n_{MgO}} = 0,2mol\\
a)\\
\% {m_{Mg}} = \dfrac{{0,15 \times 24}}{{11,6}} \times 100\%  = 31,03\% \\
\% {m_{MgO}} = 100\%  - 31,03\%  = 68,97\% \\
b)\\
{n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,35mol\\
 \to {m_{MgC{l_2}}} = 33,25g\\
{n_{{H_2}}} = {n_{Mg}} = 0,15mol \to {m_{{H_2}}} = 0,3g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 411,3g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{33,25}}{{411,3}} \times 100\%  = 8,08\% 
\end{array}\)

Đặt `n_(Mg)=x(mol)`, `n_(MgO)=y(mol)`

`→24x+40y=11.6 (*)`

`PTHH: Mg+2HCl→MgCl_2+H_2 (1)`

`MgO+2HCl→MgCl_2+H_2O (2)`

`NaOH+HCl→NaCl+H_2O (3)`

`n_(HCl)=(400·7.3%)/(100%·36.5)=0.8(mol)`

`m_(NaOH)=(400·4%)/(100%·40)=0.1(mol)`

`Theopt: n_(HCl)(3)=n_(NaOH)=0.1(mol)`

`→∑n_(HCl)(1)+(2)=0.8-0.1=0.7(mol)(**)`

Từ `(*)(**)→x=0.15(mol), y=0.2(mol)`

`→m_(Mg)=0.15·24=3.6(g)`

`→%m_(Mg)=3.6/11.6·100%=31.03%`

`→%m_(MgO)=100%-31.03%=68.97%`

`b,` Dd thu đc chứa `MgCl_2`

Ta có:

`∑n_(MgCl_2)=n_(Mg)+n_(MgO)=0.15+0.2=0.35(mol)`

`→m_(MgCl_2)=0.35·95=33.25`

`n_(H_2)=n_(Mg)=0.15(mol)`

`m_(dd)=m_(hh)+m_(HCl)-m_(H_2)`

`=11.6+400-0.15·2=411.3(g)`

`C%(ddMgCl_2)=33.25/411.3·100%=8.08%`