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Câu 13: a. `-` `Fe + 2HCl -> FeCl_2 + H_2`

b. `-` $n_{Fe}$ = $\dfrac{5,6}{56}$ = `0,1` `(mol)`

`-` Theo phương trình $n_{H_2}$ = $n_{Fe}$ = `0,1` `(mol)` 

`->` $V_{H_2(đktc)}$ = `0,1 × 22,4` = `2,24` `(l)`

Câu 14: 1. `2KClO_3 \overset{t^o}\to 2KCl + 3O_2`

2. Sắt + khí oxi `\overset{t^o}\to` Sắt (II) oxit

3. `4P + 5O_2 \overset{t^o}\to 2P_2O_5`

4. Nhôm + oxi `\overset{t^o}\to` Nhôm oxit

5. `Al + O_2 \overset{t^o}\to Al_2O_3`

6. `4H_2 + Fe_3O_4 \overset{t^o}\to 3Fe + 4H_2O`

7. `4P + 5O_2 \overset{t^o}\to 2P_2O_5`

8. `2KClO_3 \overset{t^o}\to 2KCl + 3O_2`

9. `S + O_2 \overset{t^o}\to SO_2`

10. `PbO + H_2 -> Pb + H_2O`

Em tham khảo nha :

\(\begin{array}{l}
13)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{5,6}}{{56}} = 0,1mol\\
{n_{{H_2}}} = {n_{Fe}} = 0,1mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l\\
14)\\
1)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
2)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
3)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
4)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
5)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
6)\\
F{e_3}{O_4} + 4{H_2} \xrightarrow{t^0} 3Fe + 4{H_2}O\\
7)\\
4P + 5{O_2} \xrightarrow{t^0} 2{P_2}{O_5}\\
8)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
9)\\
S + {O_2} \xrightarrow{t^0} S{O_2}\\
10)\\
PbO + {H_2} \xrightarrow{t^0} Pb + {H_2}O
\end{array}\)