Em tham khảo nha:
\(\begin{array}{l}
VD4)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{HCl}} = 0,2 \times 1 = 0,2\,mol\\
{n_{Fe}} = \dfrac{{{n_{HCl}}}}{2} = 0,1\,mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
b)\\
{n_{{H_2}}} = {n_{Fe}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
VD5)\\
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{HCl}} = \dfrac{{150 \times 14,6\% }}{{36,5}} = 0,6\,mol\\
{n_{Mg}} = \dfrac{{{n_{HCl}}}}{2} = 0,3\,mol\\
{m_{Mg}} = 0,3 \times 24 = 7,2g\\
b)\\
{n_{{H_2}}} = {n_{Mg}} = 0,3\,mol\\
{V_{{H_2}}} = 0,3 \times 22,4 = 6,72l\\
VD6)\\
A + 2HCl \to AC{l_2} + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_A} = {n_{{H_2}}} = 0,2\,mol\\
{M_A} = \dfrac{{4,8}}{{0,2}} = 24g/mol \Rightarrow A:Mg
\end{array}\)
