Đáp án:

Giải thích các bước giải:

$n_{Al}$= $\frac{4,05}{27}$= $0,15(mol)$

$n_{O_2}$= $\frac{2,016}{22,4}$= $0,09(mol)$

$Pt: 4Al + 3O_2$ $\xrightarrow{t^o}$ $2Al_2O_3$

$\text{Tỉ lệ: $\dfrac{0,15}{4}$ > $\dfrac{0,09}{3}$}$

$\rightarrow$ $\text{Al dư, $O_2$ hết.}$

$m_{Al(dư)}$= $(0,15-0,12).27 = 0,81(g)$

$m_{Al_2O_3}$= $0,06.102= 6,12(g)$

`n_(Al)=(m_(Al))/(M_(Al))=(4,05)/(27)=0,15` (mol)

`n_(O_2)=(V_(O_2))/(22,4)=(2,016)/(22,4)=0,09` (mol)

PTHH:

`4Al+3O_2 → 2Al_2 O_3`

  4       3                 2

`(n_(Al))/4=0,0375` `>` `(n_(O_2))/3=0,03`

`=>Al` đủ

`n_(Al_2 O_3)=(0,09 . 2)/3=0,06` (mol)

`=>m_(Al_2 O_3)=n_(Al_2 O_3). M_(Al_2 O_3)=0,06.102=6,12` (g)