Đáp án:

Xác định số oxi hoá:

$\rm \mathop{Zn}\limits^{0}+H_2\mathop{S}\limits^{+6}O_4 \to \mathop{Zn}\limits^{+2}SO_4+ H_2\mathop{S}\limits^{-2}+H_2O$

Thăng bằng electron:

$\begin{array}{c|cc} \rm{\times \;4}&\rm{\mathop{Mg}\limits^{0}\longrightarrow\mathop{Mg}\limits^{+2}+2e}\\\rm {\times\;1}&\rm{\mathop{S}\limits^{+6}+8e\longrightarrow\mathop{S}\limits^{-2}}\\\end{array}$

Cân bằng phương trình:

$\rm 4Zn + 5H_2SO_4\to 4ZnSO_4 + H_2S + 4H_2O$

Xác định số oxi hoá:

$\rm K\mathop{Mn}\limits^{+7}O_4+ H\mathop{Cl}\limits^{-1} \to \mathop{Cl_2}\limits^{0}+\mathop{Mn}\limits^{+2}Cl_2+KCl+H_2O$

Thăng bằng electron:

$\begin{array}{c|cc} \rm{\times \;5}&\rm{2\mathop{Cl}\limits^{-1}\longrightarrow\mathop{Cl_2}\limits^{0}+2e}\\\rm {\times\;1}&\rm{\mathop{Mn}\limits^{+7}+5e\longrightarrow\mathop{Mn}\limits^{+2}}\\\end{array}$

Cân bằng phương trình:

$\rm 2 KMnO_4 +16 HCl\to 5 Cl_2 + 2MnCl_2 + 2KCl +8 H_2O$

$\rm \mathop{Zn}\limits^{0}+H_2\mathop{S}\limits^{+6}O_4 \to \mathop{Zn}\limits^{+2}SO_4+ H_2\mathop{S}\limits^{-2}+H_2O$

$\begin{array}{c|cc} \rm{\times \;4}&\rm{\mathop{Zn}\limits^{0}\longrightarrow\mathop{Zn}\limits^{+2}+2e}\\\rm {\times\;1}&\rm{\mathop{S}\limits^{+6}+8e\longrightarrow\mathop{S}\limits^{-2}}\\\end{array}$

`=>` $\rm 4Zn+5H_2SO_4 \to 4ZnSO_4+H_2S+4H_2O$

`-----`

$\rm K\mathop{Mn}\limits^{+7}O_4+ H\mathop{Cl}\limits^{-1} \to \mathop{Cl_2}\limits^{0}+\mathop{Mn}\limits^{+2}Cl_2+KCl+H_2O$

$\begin{array}{c|cc} \rm{\times \;5}&\rm{\mathop{2Cl}\limits^{-1}\longrightarrow\mathop{Cl_2}\limits^{0}+2e}\\\rm {\times\;2}&\rm{\mathop{Mn}\limits^{+7}+5e\longrightarrow\mathop{Mn}\limits^{+2}}\\\end{array}$

`=>` $\rm 2KMnO_4+16HCl \to 5Cl_2+2MnCl_2+2KCl+8H_2O$