$\rm PTHH: 2KMnO_4 + 16HCl \xrightarrow{} 2KCl +2MnCl_2 +5Cl_2 +8H_2O$ 

$n_{KMnO_4}=\dfrac{3,16}{158}=0,02(mol)$ 

theo pthh 

$n_{Cl_2}=\dfrac{5}{2}n_{KMnO_4}=0,05(mol)$ 

`->`$V_{Cl_2}=0,05.22,4=1,12(l)$ 

`@`

theo pthh 

$n_{KMnO_4}=\dfrac{2}{5}n_{Cl_2}=0,2(mol)$ 

`->` $m_{KMnO_4}=0,2.158=31,6(g)$ 

mà $H=80\%$ 

`->`$m_{KMnO_4(tt)}=\dfrac{31,6.100}{80}=39,5(g)$ 

Đáp án:

$\rm 1.\\n_{KMnO_4}=\dfrac{3,16}{158}=0,02\ (mol)$

Phản ứng xảy ra:

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

Theo phản ứng:

$\rm n_{Cl_2}=\dfrac{5}{2}n_{KMnO_4}=0,05\ (mol)$

$\rm V_{Cl_2}=0,05×22,4=1,12\ (L)$

$2.$

Theo phản ứng:

$\rm n_{KMnO_4}=\dfrac{2}{5}n_{Cl_2}=0,2\ (mol)$

Khối lượng $\rm KMnO_4$ dùng:

$\rm m_{KMnO_4}=\dfrac{0,2×158}{80\%}=39,5\ (g)$