Đáp án:

`-` `C_{M_{NaCl}}=C_{M_{NaClO}}=1,5\ (M).`

Giải thích các bước giải:

`-` `n_{KMnO_4}=\frac{15,8}{158}=0,1\ (mol).`

`-` `n_{NaOH}=0,1\times 3=0,3\ (mol).`

Phương trình hóa học:

`2KMnO_4 + 16HCl \overset{t^o}\to 2KCl + 2MnCl_2 + 5Cl_2\uparrow + 8H_2O` (1)

`Cl_2 + 2NaOH \to NaCl + NaClO + H_2O` (2)

`-` Theo phương trình (1): `n_{Cl_2}=\frac{5}{2}n_{KMnO_4}=0,25\ (mol).`

`-` Tỉ lệ: `n_{Cl_2}:n_{NaOH}=\frac{0,25}{1}>\frac{0,3}{2}`

`\to Cl_2` dư, `NaOH` hết.

`-` Theo phương trình (2): `n_{NaCl}=n_{NaClO}=\frac{1}{2}n_{NaOH}=0,15\ (mol).`

`\to C_{M_{NaCl}}=C_{M_{NaClO}}=\frac{0,15}{0,1}=1,5\ (M).`

\(\boxed{\text{LOVE TEAM}}\)

`2KMnO_4+16HCl->2KCl+2MnCl_2+5Cl_2↑+8H_2O`

`Cl_2+2NaOH->NaClO+NaCl+H_2O`

`n_(KMnO_4)=(15,8)/(39+55+16.4)=0,1(mol)`

`=>n_(Cl_2)=0,25(mol)`

`n_(NaOH)=0,1.3=0,3(mol)`

`(0,3)/2<(0,25)/1`

`=>Cl_2` dư.

`(n_(NaOH))/2=n_(NaClO)=n_(NaCl)=0,15(mol)`

`=>C_(M_(NaClO))=C_(M_(NaCl))=(0,15)/(0,1)=1,5M`