$8/B$
$II.$
$1/$
$Na2O=$Natri oxit
$CO=$cacbon monoxit
$FeO=$sắt (II) oxit
$SO2=$ lưu huỳnh đioxit
$Al2O3=$nhôm oxit
$P2O5=$điphotpho pentaoxit
$2/$
$a. 4P+5O 2 → 2P 2 O 5$
$b. 4Al+3O 2 → 2Al 2 O 3$
$c. 2KClO 3 → 2KCl+3O 2 .$
$d. CaCO 3 → CaO + CO 2 .$
$e. CH 4 +2O 2 → CO 2 +2H 2 O$
$3/$
$a/$
$4Fe + 3O2 → 2Fe2O3$
$b/$
$nFe=25,2/56=0,45mol$
$⇒nO2=3/4nFe=3/4.0,45=0,3375mol$
$⇒V_{O_{2}}=0,3375.22,4=7,56l$
$c/$
$pthh:$
$2KClO3→2KCl+3O2$
$theo$ $pt :$
$nKClO3=2/3.nO2=2/3.0,3375=0,225mol$
$⇒mKClO3=0,225.122,5=27,5625g$