Em tham khảo nha:

\(\begin{array}{l}
6)\\
a)\\
3Fe + 2{O_2} \xrightarrow{t^0} F{e_3}{O_4}\\
{n_{Fe}} = \dfrac{{16,8}}{{56}} = 0,3\,mol\\
{n_{{O_2}}} = 0,3 \times \dfrac{2}{3} = 0,2\,mol\\
{V_{{O_2}}} = 0,2 \times 22,4 = 4,48l\\
b)\\
2KCl{O_3} \xrightarrow{t^0} 2KCl + 3{O_2}\\
{n_{KCl{O_3}}} = 0,2 \times \dfrac{2}{3} = \dfrac{2}{{15}}\,mol\\
{m_{KCl{O_3}}} = \dfrac{2}{{15}} \times 122,5 = 16,33g\\
7)\\
2A + 2nHCl \to 2AC{l_n} + n{H_2}\\
{n_{{H_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6\,mol\\
{n_A} = 0,6 \times \dfrac{2}{n} = \dfrac{{1,2}}{n}\,mol\\
{M_A} = \dfrac{{10,8}}{{\dfrac{{1,2}}{n}}} = 9n(g/mol)\\
n = 1 \Rightarrow {M_A} = 9g/mol\\
n = 2 \Rightarrow {M_A} = 18g/mol\\
n = 3 \Rightarrow {M_A} = 27g/mol \Rightarrow A:Al
\end{array}\)

Đáp án + Giải thích các bước giải:

Câu VI

 `a)` 

`3Fe + 2O_2`$\xrightarrow{t°}$`Fe_3O_4`

`n_{Fe} = {16,8}/{56}= 0,3(mol)`

`n_{O_2} = 2/3n_{Fe}`

`= 2/3.  0,3= 0,2(mol)`

`V_{O_2} = 0,2 .22,4= 4,48(l)`

`b)`

`2KClO_3`$\xrightarrow{t°}$`2KCl+ 3O_2`

`n_{KClO_3} = 2/3n_{O_2}`

`= 2/3.  0,2= 2/{15}(mol)`

`m_{KClO_3}= 2/{15}. 122,5= 16,33(g)`

Câu VII: 

`2A + 2nHCl→ 2ACl_n + nH_2`

`n_{H_2} = {13,44}/{22,4} = 0,6(mol)`

`n_A= {2n_{H_2}}/n`

`= {0,6.2}/n= {1,2}/n(mol)`

`M_A= {10,8}/{{1,2}/n} = 9n`$(g/mol)$

`n= 3→ A= 27`$(g/mol)$

`→ A:Al`