Đáp án + Giải thích các bước giải:

$a\big)$

$n_{Al}=\frac{21,6}{27}=0,8(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

Theo PT: $n_{H_2}=\frac{3}{2}n_{Al}=1,2(mol)$

$\to V_{H_2(đktc)}=1,2.22,4=26,88(l)$

$b\big)$

Theo PT: $n_{HCl}=2n_{H_2}=2,4(mol)$

$\to m_{HCl}=2,4.36,5=87,6(g)$

$\to m_{dd\,HCl}=\frac{87,6}{14,6\%}=600(g)$

$c\big)$

Theo PT: $n_{AlCl_3}=n_{Al}=0,8(mol)$

Ta có $m_{dd\,spu}=21,6+600-1,2.2=619,2(g)$

$\to C\%_{AlCl_3}=\frac{0,8.133,5}{619,2}.100\%\approx 17,25\%$

Đáp án+ Giải thích các bước giải:

`a)`

`PT : 2Al + 6HCl -> 2AlCl_3 + 3H_2`

`n_{Al} = m/M = {21,6}/27 = 0,8(mol)`

Theo pt : `n_{H_2} = 3/2.``n_{Al} = 3/2.``0,8 = 1,2(mol)`

`-> V_{H_2} = n.22,4 = 1,2.22,4 = 26,88(l)`

`b)`

Theo pt : `n_{HCl} = 3n_{Al} = 3.0,8 = 2,4(mol)`

`-> m_{HCl} = n.M = 2,4.36,5 = 87,6(g)`

`c)`

`m_{ddHCl} = {87,6.100%}/{14,6%} = 600(g)`

`m_{dd(sau)} = 21,6+600 - 1,2.2 = 619,2(g)`

Theo pt : `n_{AlCl_3} = n_{Al} = 0,8(mol)`

`-> C%`$_{dd FeCl_2}$ `= \frac{0,8.133,5.100%}{619,2} \approx 17,25%`

$\text{#Cyrus}$