$BTKL$`->`$n_{O_2}=\dfrac{23.15-19.15}{32}=0.125(mol)$

$n_{Cl_2}=\dfrac{4.4622}{24.79}=0.18(mol)$

$n_{KMnO_4}=a(mol);n_{KClO_3}=b(mol)$

`->`$158a+122.5b=23.15$ $(1)$

$\rm \mathop{Mn}\limits^{+7}+5e \to \mathop{Mn}\limits^{+2}\, \, || \,\, \mathop{2Cl}\limits^{-1}\to \mathop{Cl_2}\limits^{0}+2e$

$\rm \mathop{Cl}\limits^{+5}+6e \to \mathop{Cl}\limits^{-1} \,\, || \,\, \mathop{2O}\limits^{-2} \to \mathop{O_2}\limits^{0}+4e$

Bảo toàn $e:$

$5a+6b=4n_{O_2}+2n_{Cl_2}=0.86$ $(2)$

Từ $(1),(2)$ suy ra: $a=0.1;b=0.06$

$\%m_{KMnO_4}=68.25\%$

$\%m_{KClO_3}=31.75\%$

$\rm \mathop{Mn}\limits^{+7}+5e \to \mathop{Mn}\limits^{+4} \,\, || \,\, \mathop{2O}\limits^{-2} \to \mathop{O_2}\limits^{0}+4e$

$\rm \mathop{Mn}\limits^{+7}+1e \to \mathop{Mn}\limits^{+6} \,\, ||$

Bảo toàn $e:$

$6 \times 0.06+4 \times (0.1-n_{KMnO_4,dư})=4n_{O_2}=0.5$

`=>`$n_{KMnO_4,dư}=0.065(mol)$

`=>`$\%m_{KMnO_4,\text{bị nhiệt phân}}=\dfrac{0.1-0.065}{0.1} \times 100\%=35\%$

$b)$

$n_{AgCl}=0.4(mol)$

Bảo toàn nguyên tố $Cl:$

$n_{KClO_3}+n_{HCl}=n_{AgCl}+2n_{Cl_2}$

`=>`$V =\dfrac{0.4+2 \times 0.18-0.06}{1.25 \times 10^{-3}}=560(mL)$