Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
b)\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
hh:Zn(a\,mol),Fe(b\,mol)\\
\left\{ \begin{array}{l}
65a + 56b = 12,1\\
a + b = 0,2
\end{array} \right.\\
\Rightarrow a = b = 0,1\\
\% {m_{Fe}} = \dfrac{{0,1 \times 56}}{{12,1}} \times 100\% = 46,28\% \\
\% {m_{Zn}} = 100 - 46,28 = 53,72\% \\
c)\\
{n_{HCl}} = 2{n_{{H_2}}} = 0,4\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,4}}{1} = 0,4l\\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1\,mol\\
{n_{ZnC{l_2}}} = {n_{Zn}} = 0,1\,mol\\
{C_M}FeC{l_2} = {C_M}ZnC{l_2} = \dfrac{{0,1}}{{0,4}} = 0,25M\\
2)\\
a)\\
2KMn{O_4} + 16HCl \to 2KCl + 2MnC{l_2} + 5C{l_2} + 8{H_2}O\\
C{l_2} + 2Na \to 2NaCl\\
2NaCl + 2{H_2}O \to 2NaOH + C{l_2} + {H_2}\\
C{l_2} + {H_2} \to 2HCl\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
CuC{l_2} + 2NaOH \to Cu{(OH)_2} + 2NaCl\\
b)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O\\
C{l_2} + {H_2} \to 2HCl\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
FeC{l_2} + 2NaOH \to Fe{(OH)_2} + 2NaCl\\
2Fe + 3C{l_2} \to 2FeC{l_3}
\end{array}\)
