- Chọn B.

- $2KCl{{O}_{3}}\xrightarrow{{{t}^{o}}}2KCl+3{{O}_{2}}$                    $2KMn{{O}_{4}}\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$

    2(mol)                      3 (mol)              2 (mol)                                       1 (mol)

  $x\left( mol \right)$               $\to \frac{3}{2}x$ (mol)                $y\left( mol \right)$                 $\to $                $\frac{1}{2}$y(mol)

- Theo bài ra ta có hệ PT:

$\left\{ \begin{align} & 122,5x+158y=205,05 \\  & \frac{2}{3}x+\frac{1}{2}y=\frac{36,96}{22,4}=1,65 \\ \end{align} \right.$

$\Leftrightarrow \left\{ \begin{align}  & x=0,9\left( mol \right) \\  & y=0,6(mol) \\\end{align} \right.$

Khối lượng của muối $KMn{{O}_{4}}$có trong hỗn hợp là: ${{m}_{KMn{{O}_{4}}}}=0,6\times 158=94,8\left( gam \right)$