Đáp án:
PTHH :
$\text{ 4Al + $3O_{2}$ $\to$ $2Al_{2}O_{3}$}$
$n_{Al} = 2,7 : 27 = 0,1$ mol
$⇒n_{Al_{2}O_{3}} = \dfrac{0,1.2}{4} = 0,05$ mol
$⇒m_{Al_{2}O_{3}} =0,05 . 102= 5,1$ g
$⇒n_{O_{2}} = \dfrac{0,1 .3}{4}=0,075$ mol
$⇒V_{O_{2}} = 0,075 . 22,4 = 1,68$ l
Ta có $5V_{O_{2}} = V_{kk}$
$⇒V_{kk} = 5 . 1,68 = 8,4$ l