Đáp án:

 a)

\(\begin{array}{l}
\% Mg = 13,04\% \\
\% MgO = 86,96\% \\
b)\\
{m_{NaOH}} = 4g
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
 \Rightarrow {n_{Mg}} = 0,05mol\\
{m_{Mg}} = n \times M = 0,05 \times 24 = 1,2g\\
{m_{MgO}} = m - {m_{Mg}} = 9,2 - 1,2 = 8g\\
\% Mg = \dfrac{{{m_{Mg}}}}{m} \times 100\%  = \dfrac{{1,2}}{{9,2}} \times 100\%  = 13,04\% \\
\% MgO = 100 - 13,04 = 86,96\% \\
b)\\
{n_{MgO}} = \dfrac{m}{M} = \dfrac{8}{{40}} = 0,2mol\\
{n_{HCl}} = V \times {C_M} = 0,2 \times 3 = 0,6mol\\
{n_{HC{l_d}}} = {n_{HCl}} - 2{n_{Mg}} - 2{n_{MgO}}\\
\,\,\,\,\,\,\,\,\,\, = 0,6 - 2 \times 0,05 - 2 \times 0,2 = 0,1mol\\
HCl + NaOH \to NaCl + {H_2}O\\
 \Rightarrow {n_{NaOH}} = {n_{NaC{l_d}}} = 0,1mol\\
{m_{NaOH}} = n \times M = 0,1 \times 40 = 4g
\end{array}\)

`a,`

`n_(H_2)=\frac{1,12}{22,4}=0,05(mol)`

`n_(HCl)=0,3.2=0,6(mol)`

`Mg+2HCl->MgCl_2+H_2`

`0,05`      `0,1`                    `0,05`

`MgO+2HCl->MgCl_2+H_2O`

`0,2`         `0,4`

`=>m_(Mg)=0,05.24=1,2(g)`

`=>m_(MgO)=9,2-1,2=8(g)`

`=>n_(MgO)=\frac{8}{40}=0,2(mol)`

`=>%m_(Mg)=\frac{1,2}{9,2}=13,04%`

`%m_(MgO)=100-13,04=86,96%`

`b,`

`HCl+NaOH->NaCl+H_2O`

`0,1`           `0,1`

`=>m_(NaOH)=0,1.40=4(g)`