Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
MgO + 2HCl \to MgC{l_2} + {H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mo\\
{n_{Mg}} = {n_{{H_2}}} = 0,05mol\\
{m_{Mg}} = 0,05 \times 24 = 1,2g\\
{m_{MgO}} = 5,2 - 1,2 = 4g\\
{n_{MgO}} = \dfrac{4}{{40}} = 0,1mol\\
{n_{HC{l_{pu}}}} = 2{n_{Mg}} + 2{n_{MgO}} = 0,3mol\\
{n_{HCl}} = \dfrac{{0,3 \times 125}}{{100}} = 0,375mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,375}}{{0,5}} = 0,75M\\
c)\\
2NaOH + MgC{l_2} \to Mg{(OH)_2} + 2NaCl\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{HC{l_d}}} = 0,375 - 0,3 = 0,075mol\\
{n_{MgC{l_2}}} = {n_{Mg}} + {n_{MgO}} = 0,15mol\\
{n_{NaOH}} = 2{n_{MgC{l_2}}} + {n_{HCl}} = 0,375mol\\
{V_{NaOH}} = \dfrac{{0,375}}{2} = 0,1875l = 187,5ml
\end{array}\)