Em tham khảo nha :

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{Mg}} = {n_{{H_2}}} = 0,2mol\\
{m_{Mg}} = 0,2 \times 24 = 4,8g\\
{n_{HCl}} = \dfrac{{36,5}}{{36,5}} = 1mol\\
{n_{MgO}} = \dfrac{{1 - 0,2 \times 2}}{2} = 0,3mol\\
{m_{MgO}} = 0,3 \times 40 = 12g\\
\% Mg = \dfrac{{4,8}}{{12 + 4,8}} \times 100\%  = 28,6\% \\
\% MgO = 100 - 28,6 = 71,4\% \\
{n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,2 + 0,3 = 0,5mol\\
{m_{MgC{l_2}}} = 0,5 \times 95 = 47,5g
\end{array}\)