Em tham khảo nha:

\(\begin{array}{l}
4)\\
4Al + 3{O_2} \xrightarrow{t^0} 2A{l_2}{O_3}\\
a)\\
{n_{A{l_2}{O_3}}} = \dfrac{{40,8}}{{102}} = 0,4\,mol\\
{n_{Al}} = 2{n_{A{l_2}{O_3}}} = 0,8\,mol\\
{m_{Al}} = 0,8 \times 27 = 21,6g\\
b)\\
{n_{{O_2}}} = 0,4 \times \frac{3}{2} = 0,6\,mol\\
{V_{{O_2}}} = 0,6 \times 22,4 = 13,44l\\
{V_{kk}} = 13,44 \times 5 = 67,2l\\
5)\\
C + {O_2} \xrightarrow{t^0}C{O_2}\\
{n_C} = \dfrac{{18000}}{{12}} = 1500\,mol\\
{n_{{O_2}}} = \dfrac{{24000}}{{32}} = 750\,mol\\
\text{ Lập tỉ lệ }:{n_C} > {n_{{O_2}}}(1500 > 750)\\
 \Rightarrow \text{ C dư } \\
{n_C} \text{ dư }= 1500 - 750 = 750\,mol\\
{m_C} \text{ dư } = 750 \times 12 = 9000g = 9kg\\
{n_{C{O_2}}} = {n_{{O_2}}} = 750\,mol\\
{m_{C{O_2}}} = 750 \times 44 = 33000g = 33kg
\end{array}\)

Câu 4:

nAl₂O₃ = $\frac{40,8}{27}$ = 1,5 (mol)

PTHH: 2Al + 3O₂ → 2Al₂O₃

           1,5     2,25     1,5      (mol)

mAl = 1,5 . 2 . 27 = 81 (g)

VO₂ = 2,25 . 22,4 = 50,4 (l)

Vkk = $\frac{50,4}{20%}$ = 252 (l)