- Chọn A.

- Ta có: ${{n}_{{{O}_{2}}}}=\frac{11,2}{22,4}=0,5\left( mol \right)$.

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

               2 (mol)                                          1 (mol)

               x (mol)                                          0,5 (mol)

${{n}_{KMn{{O}_{4}}}}=x=\frac{0,5\times 2}{1}=1\left( mol \right)$

$\to {{m}_{KMn{{O}_{4}}}}=1\times 158=158\left( gam \right)$.