$1/C$
$2/B$
$3/A$
$4/D$
Bài 1:
$Al2O3:$ nhôm oxit
$Fe2O3:$ sắt(III) oxit
$P2O3$: điphotpho trioxit
$H2O$ : nước
Bài 2:
a/$3Fe + 2O2$ ------> $2Fe3O4$
b/2KNO3→2KNO2+O2
c/ 2Al + Cl2 → 2AlCl3
$bài 3 :$
$a/$
4Fe + 3O2 → 2Fe2O3
$b/$
$nFe = 2,25 (mol)$
$theo$ $pt: $
$nO2=3/4nFe=3/4.2,25=1,6875(mol)$
$→VO2=1,6875.22,4=37,8 (l)$
$c/$
$pthh : $
2KClO3→2KCl+3O2
$theo$ $pt :$
$nKClO3=2/3.nO2=2/3.1,6875=1,125(mol)$
$→mKClO3=1,125.122,5=137,8125 (g)$