Em tham khảo nha:

\(\begin{array}{l}
VD5)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{HCl}} = \dfrac{{150 \times 14,6\% }}{{36,5}} = 0,6\,mol\\
{n_{Mg}} = \dfrac{{{n_{HCl}}}}{2} = 0,3\,mol\\
{m_{Mg}} = 0,3 \times 24 = 7,2g\\
b)\\
{n_{{H_2}}} = {n_{Mg}} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
VD6)\\
A + 2HCl \to AC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_A} = {n_{{H_2}}} = 0,2\,mol\\
{M_A} = \dfrac{{4,8}}{{0,2}} = 24g/mol \Rightarrow A:Mg\\
VD7)\\
2A + 2{H_2}O \to 2AOH + {H_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_A} = 2{n_{{H_2}}} = 0,2\,mol\\
{M_A} = \dfrac{{4,6}}{{0,2}} = 23g/mol \Rightarrow A:Natri(Na)
\end{array}\)