gọi $n_{KMnO4 pư}=x(mol)$
$⇒n_{KClO3 pư}=3x(mol)$
$PTHH:$ $2KMnO4\overset{t^{o}}{\rightarrow}K2MnO4+MnO2+O2(1)$
x 0,5x
$ 2KClO3\overset{t^{o}}{\rightarrow}2KCl+3O2(2)$
2x 3x
ta có:
$m_{hh}= 158x+122,5.2x=40,3g$
$⇒x=0,1$
$⇒n_{O2}=n_{O2(1)}+n_{O2(2)}=0,5x+3x=0,35mol$
$⇒V_{O2}=0,35.22,4=7,84l$