$\text{Đáp án+Giải thích các bước giải:}$

$\text{Câu 1:}$

$\text{a) $Fe_{3}O_{4}$ + 8HCl→$FeCl_{2}$ + $2FeCl_{3}$ + $4H_{2}O$}$

$\text{b) $N_{2}O_{5}$ + $H_{2}O$→$2HNO_{3}$}$

$\text{c) $K_{2}O$ + $H_{2}O$→2KOH}$

$\text{d) 2Fe + $6H_{2}SO_{4}$→$Fe_{2}(SO_{4})_{3}$ + $3SO_{2}$ + $6H_{2}O$}$

$\text{e) 3Mg + $2H_{3}PO_{4}$→$Mg_{3}(PO_{4})_{2}$ + $3H_{2}$}$

$\text{Câu 2:}$

$\text{$2KClO_{3}$$\xrightarrow{t^o}$$3O_{2}$ + 2KCl}$

$\text{$2O_{2}$ + 3Fe$\xrightarrow{t^o}$$Fe_{3}O_{4}$}$

$\text{$Fe_{3}O_{4}$ + $4H_{2}$$\xrightarrow{t^o}$3Fe + $4H_{2}O$}$

$\text{$2H_{2}$ + $O_{2}$$\xrightarrow{t^o}$$2H_{2}O$}$

$\text{A: $O_{2}$}$

$\text{B: KCl}$

$\text{C: Fe}$

$\text{D: $H_{2}$}$

$\text{E: $H_{2}O$}$

Đáp án:

Câu 1:

a/  $Fe_{3}$$O_{4} + 8HCl → $FeCl_{2}$ + $FeCl_{3}$ +$H_{2}$O

b/  $N_{2}$$O_{5}$ + $H_{2}$O → 2$HNO_{3}$

c/  $K_{2}$O + $H_{2}$O → 2KOH

d/ 2Fe + 6$H_{2}$$SO_{4}$ → $Fe_{2}$$($SO_{4}$)_{3}$ + 3$SO_{2}$ + 6$H_{2}$O

e/ 3Mg + 2$H_{3}$$PO_{4}$ → $Mg_{3}$$($PO_{4}$)_{2}$ + 3$H_{2}$

Câu 2:

4$KClO_{3}$ → KCl + 3$KClO_{4}$

$Fe_{3}$$O_{4}$ + 4CO → 3Fe + 4$CO_{2}$

3Fe + 2$O_{2}$ → $Fe_{3}$$O_{4}$