Giải thích các bước giải:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
a)\\
{n_{{H_2}}} = 0,05mol\\
 \to {n_{Mg}} = {n_{{H_2}}} = 0,05mol \to {m_{Mg}} = 1,2g\\
 \to {m_{MgO}} = 8g \to {n_{MgO}} = 0,2mol\\
 \to \% {m_{Mg}} = \dfrac{{1,2}}{{9,2}} \times 100\%  = 13,04\% \\
 \to \% {m_{MgO}} = 100\%  - 13,04\%  = 86,96\% \\
b)\\
{n_{HCl}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,5mol \to {m_{HCl}} = 18,25g\\
 \to {m_{{\rm{dd}}HCl}} = \dfrac{{18,25}}{{14,6\% }} \times 100\%  = 125g\\
c)\\
{n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,25mol \to {m_{MgC{l_2}}} = 23,75g\\
{m_{{\rm{dd}}}} = {m_{hh}} + {m_{{\rm{dd}}HCl}} - {m_{{H_2}}} = 134,1g\\
 \to C{\% _{MgC{l_2}}} = \dfrac{{23,75}}{{134,1}} \times 100\%  = 17,71\% 
\end{array}\)

Đặt `n_(Mg)=x(mol), n_(MgO)=y(mol)`

`→24x+40y=9.2`

`Mg+2HCl→MgCl_2+H_2`

` x.........2x............x.............x(mol)`

`MgO+2HCl→MgCl_2+H_2O`

`   y...........2y...........y..............y(mol)`

`n_(Mg)=x=n_(H_2)=1.12/22.4=0.05(mol)`

`→24·0.05+40·y=9.2`

`→y=0.2(mol)`

`a, %m_(Mg)=(0.05·24)/9.2·100%≈13%`

`%m_(MgO)=(0.2·40)/9.2·100%≈87%`

`b, n_(HCl)=2·0.05+2·0.2=0.5(mol)`

`m=m_(ddHCl)=(0.5·36.5·100%)/(14.6%)=125(g)`

`c,` Dd sau pư chứa `MgCl_2`

`∑n_(MgCl_2)=x+y=0.05+0.2=0.25(mol)`

Theo ĐLBTKL:

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

`=9.2+125-0.05·2=134.1(g)`

`C%(ddMgCl_2)=(0.25·95)/134.1·100%≈17.71%`