- Chọn C.

-             $2KCl{{O}_{3}}\xrightarrow{{{t}^{o}}}2KCl+3{{O}_{2}}$                       

                2(mol)                      3 (mol)

            $\frac{m}{122,5}$ (mol)                 $\frac{{{V}_{1}}}{22,4}$  (mol)

$\frac{{{V}_{1}}}{22,4}=\frac{m\times 3}{122,5\times 2}=\frac{m}{81,67}\left( mol \right)$$\left( 1 \right)$

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

 2 (mol)                                          1 (mol)

  $\frac{m}{148}$ (mol)                                      $\frac{{{V}_{2}}}{22,4}$ (mol)

$\frac{{{V}_{2}}}{22,4}=\frac{m\times 1}{158\times 2}=\frac{m}{316}\left( mol \right)$$\left( 2 \right)$

- Từ $\left( 1 \right)$và $\left( 2 \right)$ta suy ra:  ${{V}_{1}}>{{V}_{2}}.$