Em tham khảo nha :
\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{MgC{l_2}}} = \dfrac{{4,25}}{{95}} = 0,045mol\\
{n_{Mg}} = {n_{MgC{l_2}}} = 0,045mol\\
{m_{Mg}} = 0,045 \times 24 = 1,08g\\
b)\\
{n_{{H_2}}} = {n_{MgC{l_2}}} = 0,045mol\\
{V_{{H_2}}} = 0,045 \times 22,4 = 1,008l\\
c)\\
{n_{HCl}} = 2{n_{MgC{l_2}}} = 0,09mol\\
{m_{HCl}} = 0,09 \times 36,5 = 3,285g\\
C{\% _{HCl}} = \dfrac{{3,285}}{{100}} \times 100\% = 3,285\% \\
d)\\
{m_{ddspu}} = 1,08 + 100 - 0,09 = 100,99g\\
C{\% _{MgC{l_2}}} = \dfrac{{4,25}}{{100,99}} \times 100\% = 4.21\% \\
e)\\
MgC{l_2} + 2NaOH \to Mg{(OH)_2} + 2NaCl\\
{n_{Mg{{(OH)}_{ 2}}}} = {n_{MgC{l_2}}} = 0,045mol\\
{m_{Mg{{(OH)}_2}}} = 0,045 \times 58 = 2,61g
\end{array}\)