$n_{KMnO_4}=\dfrac{31,6}{158}=0,2(mol)$
$n_{O_2}=\dfrac{2,016}{22,4}=0,09(mol)$
$\text{PTHH:}$
$2KMnO_4 \xrightarrow{t^0} K_2MnO_4+MnO_2+O_2$
$\text{Do:} \dfrac{0,2}{2}>\dfrac{0,09}{1}$
$\to \text{Kê theo mol của} O_2$
$\to n_{KMnO_4}=0,09.2=0,18(mol)$
$\to m_{KMnO_4}=0,18.158=28,44g$
$\text{Hiệu suất:}$
%$H=\dfrac{m_{KMnO_4\text{..pứ}}}{m_{KMnO_4\text{..bđ}}}=\dfrac{28,44}{31,6}.100$%=$90$%