- 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)

               $\frac{a}{122,5}\left( mol \right)$               x(mol)                $\frac{b}{158}\left( mol \right)$                                   x(mol)

$x=\frac{a\times 3}{122,5\times 2}=\frac{3a}{245}\left( mol \right)$                                      $x=\frac{b\times 1}{158\times 2}=\frac{b}{316}\left( mol \right)$

Suy ra $x=\frac{3a}{245}=\frac{b}{316}\to \frac{a}{b}=0,258$.