Em tham khảo nha :

\(\begin{array}{l}
a)\\
2KMn{O_4} + 16HCl \to 5C{l_2} + 2KCl + 2MnC{l_2} + 8{H_2}O\\
{n_{C{l_2}}} = \dfrac{V}{{22,4}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{n_{KMn{O_4}}} = \dfrac{2}{5}{n_{C{l_2}}} = 0,08mol\\
{m_{KMn{O_4}}} = 0,08 \times 158 = 12,64g\\
b)\\
Mn{O_2} + 4HCl \to MnC{l_2} + C{l_2} + 2{H_2}O(1)\\
2KMn{O_4} + 16HCl \to 5C{l_2} + 2KCl + 2MnC{l_2} + 8{H_2}O(2)\\
2{K_2}Cr{O_7} + 28HCl \to 4KCl + 2CrC{l_3} + 9C{l_2} + 14{H_2}O(3)\\
KCl{O_3} + 6HCl \to 3C{l_2} + 3{H_2}O + KCl(4)\\
{n_{Mn{O_2}}} = {n_{KMn{O_4}}} = {n_{{K_2}Cr{O_7}}} = {n_{KCl{O_3}}} = a\,mol\\
{n_{C{l_2}(1)}} = {n_{Mn{O_2}}} = a\,mol\\
{n_{C{l_2}(2)}} = \dfrac{5}{2}{n_{KMn{O_4}}} = 2,5a\,mol\\
{n_{C{l_2}(3)}} = \dfrac{9}{2}{n_{{K_2}Cr{O_7}}} = 4,5a\,mol\\
{n_{Cl(4)}} = 3{n_{KCl{O_3}}} = 3a\,mol\\
 \Rightarrow {K_2}Cr{O_7}\text{ tác dụng với HCl đặc dư thu được nhiều khí $Cl_2$ nhất}
\end{array}\)