$n_{KMnO_4}=500/158=3,165mol$
$2KMnO_4\overset{t^o}\to K_2MnO_4+MnO_2+O_2$
a.Theo pt :
$n_{K_2MnO_4}=n_{MnO_2}=1/2.n_{KMnO_4}=1/2.3,165=1,5825mol$
$⇒m_{K_2MnO_4}=1,5825.197=311,7525g$
$m_{MnO_2}=1,5825.87=137,6675g$
$⇒m_{rắn}=311,7525+137,6675=449,43$
$⇒H\%=\dfrac{449,43}{466}.100\%=96,44\%$
b.Theo pt :
$n_{O_2}=1/2.n_{KMnO_4}=1/2.3,165=1,5825mol$
$⇒V_{O_2}=1,5825.22,4=35,448l$
$c.\%m_{KMnO_4}=\dfrac{311,7525}{466}.100\%≈67\%$
$\%m_{MnO_2}=100\%-67\%=33\%$