$a/$
$4P+5O2\overset{t^o}{\rightarrow}2P2O5$
$b/$
$n_{P2O5}=45,44/142=0,32mol$
Theo pt:
$n_{O2}=5/2.n_{P2O5}=5/2.0,32=0,8mol$
$⇒m_{O2}=0,8.32=25,6g$
$c/$
$2KMnO4\overset{t^o}{\rightarrow}K2MnO4+MnO2+O2$
Ta có :
$n_{KMnO4}=2.n_{O2}.80\%=2.0,8.80\%=1,28mol$
$⇒m_{KMnO4}=1,28.158=202,24g$