`\color{blue}{\text{_dqtdz1st_}}`

$n_{Al}$ `=` $\dfrac{2,7}{27}$ `=0,1` mol
$n_{HCl}$ `=` $\dfrac{73.25}{100 .36,5 }$ `=0,5` mol
`PTHH : 2Al + 6HCl -> 2AlCl_3 + 3H_2`
Xét tỉ lệ : $\dfrac{0,1}{2}$ `<` $\dfrac{0,5}{6}$ `-> Al` hết , Tính theo `Al`
Pứ :`  0,1  ->   0,3 ->     0,1 ->    0,15` mol
$V_{H_2}$ `=0,15 . 24,79 = 3,7185l`
$m_{dd sau pứ}$ `=2,7 + 73 - 0,15 . 2 = 75,4g`
`C%HCl` dư `=` $\dfrac{(0,5-0,3).36,5}{75,4}$ `.100% = 9,68%`
`C%AlCl_3` `=` $\dfrac{0,1.133,5}{75,4}$ `.100% = 17,71%`
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$n_{KClO_3}$ `=` $\dfrac{36,75}{122,5}$ `=0,3` mol
$n_{O_2}$`=` $\dfrac{9,916}{24,79}$ `=0,4` mol
`PTHH :2KClO_3` $\xrightarrow[]{t^o}$ `2KCl + 3O_2`
Theo `PTHH`
`->` `3/2`$n_{KClO_3}$ `=` $n_{O_2}$ `=3/2. 0,3 = 0,45` mol
`-> H%=` $\dfrac{0,4}{0,45}$ `.100% = 88,89%``