`color{#1be01b}{-T}color{#1be01b}{i}color{#1be01b}{t}color{#1be01b}{u}color{#1be01b}{u}color{#1be01b}{u-}color{#1be01c}`

`MnO_2+4HCl` $\longrightarrow$ `MnCl_2+Cl_2+2H_2O (1)`

`2KMnO_4+16HCl` $\longrightarrow$ `2KCl+2MnCl_2+5Cl_2+8H_2O (2)`

Theo `(1), (2): n_(Cl_2)=n_(MnO_2)+(5)/(2)n_(KMnO_4)=0,2+(5)/(2).0,3=0,95 (mol)`

`-> V=0,95.24,79=23,5505 (l)`

`n_(HCl bị o xh)=2n_(Cl_2)=2.0,95=1,9 (mol)`