Hướng dẫn:

${{n}_{KMn{{O}_{4}}}}=\frac{{{m}_{KMn{{O}_{4}}}}}{{{M}_{KMn{{O}_{4}}}}}=\frac{23,7}{158}=0,15\left( mol \right)$

$2KMn{{O}_{4}}\xrightarrow{{{t}^{0}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$

  0,15          $\to $                                    0,075   (mol)

Theo phương trình ta có: ${{n}_{{{O}_{2lt}}}}=\frac{{{n}_{KMn{{O}_{4}}}}}{2}=\frac{0,15}{2}=0,075\left( mol \right)$

Mà hiệu suất quá trình là 90% ${{n}_{{{O}_{2tt}}}}={{n}_{{{O}_{2lt}}}}.H%=0,075.90%=0,0675\left( mol \right)$

${{V}_{{{O}_{2tt}}}}=22,4.{{n}_{{{O}_{2tt}}}}=22,4.0,0675=1,512\left( l \right)$

$\Rightarrow $ Chọn đáp án B