1)

Các phản ứng xảy ra:

\(2KCl{O_3}\xrightarrow{{{t^o}}}2KCl + 3{O_2}\)

\(2KMn{O_4}\xrightarrow{{{t^o}}}{K_2}Mn{O_4} + Mn{O_2} + {O_2}\)

2)

Phản ứng xảy ra:

\(4Al + 3{O_2}\xrightarrow{{{t^o}}}2A{l_2}{O_3}\)

Ta có:

\({n_{Al}} = \frac{{10,8}}{{27}} = 0,4{\text{ mol}}\)

\( \to {n_{A{l_2}{O_3}}} = \frac{1}{2}{n_{Al}} = 0,2{\text{ mol}}\)

\( \to {m_{A{l_2}{O_3}}} = 0,2.(27.2 + 16.3) = 20,4{\text{ gam}}\)

\({n_{{O_2}}} = \frac{3}{4}{n_{Al}} = 0,3{\text{ mol}}\)

\( \to {V_{{O_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)

\( \to {V_{kk}} = 5{V_{{O_2}}} = 6,72.5 = 33,6{\text{ lít}}\)

`#Dark`

Hai PTHH điều chế `O_{2}` là:

`2KMnO_{4}`  $\overset{t^o}\to$   `K_{2}``MnO_{4}`  +  `MnO_{2}`    +  `O_{2}`

`2KClO_{3}` $\overset{t^o}\to$  `2KCl  +  3O_{2}`

b) PTHH:    `4Al    +   3O_{2}`  $\overset{t^o}\to$  `2Al_{2}O_{3}`

b)  `n_{Al}=10,8:65=0,4(mol)`

`⇒n_{Al3O3}=(0,4.2)/4=0,2(mol)`

`⇒m_{Al2O3}=0,2.102=20,4(g)`

c) `n_{O2}=(0,2.3)/4=0,3(mol)`

`⇒V_{O2}=22,4.0,3=6,72(l)`

`⇒V_{KK}=6,72.5=33,6(l)`