a, PTHH: `2Al + 6HCl -> 2AlCl_3 + 3H_2↑`

b, `n_{Al} = (2,7)/27 = 0,1` (mol)

Theo PTHH, `n_{H_2} = 1,5n_{Al} = 1,5 . 0,1 = 0,15` (mol)

`V_{H_2} = 22,4 . 0,15 = 3,36` (l)

c, 

C1: Theo PTHH, `n_{HCl} = 3n_{Al} = 3 . 0,1 = 0,3` (mol)

`m_{HCl} = 0,3 . (1 + 35,5) = 10,95` (g)

C2: `m_{H_2} = 0,15 . 2 = 0,3` (g)

Theo PTHH, `n_{AlCl_3} = n_{Al} = 0,1` (mol)

`m_{AlCl_3} = 0,1 . (27 + 35,5 . 3) = 13,35` (g)

Áp dụng định luật BTKL:

`m_{Al} + m_{HCl} = m_{AlCl_3} + m_{H_2}`

`-> m_{HCl} = m_{AlCl_3} + m_{H_2} - m_{Al} = 13,35 + 0,3 - 2,7 = 10,95` (g)

d, PTHH: `2H_2 + O2` $\xrightarrow{t^o}$ `2H_2 O`

Theo PTHH, `n_{O_2} = 1/2 . n_{H_2} = 1/2 . 0,15 = 0,075` (mol)

`V_{O_2} = 22,4 . 0,075 = 1,68` (l)

`V_{kk} = 5V_{O_2} = 5 . 1,68 = 8,4` (l)

$n_{Al}=\dfrac{2,7}{27}=0,1(mol)$

$a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$b,$ Theo PT: $n_{H_2}=\dfrac{3}{2}n_{Al}=0,15(mol)$

$\Rightarrow V_{H_2(đktc)}=0,15.22,4=3,36(l)$

$c,$ Cách 1:

Theo PT: $n_{HCl}=3n_{Al}=0,3(mol)$

$\Rightarrow m_{HCl}=0,3.36,5=10,95(g)$

Cách 2:

Theo PT: $n_{AlCl_3}=n_{Al}=0,1(mol)$

$\Rightarrow m_{AlCl_3}=0,1.133,5=13,35(g)$

Áp dụng định luật BTKL:

$m_{Al}+m_{HCl}=m_{AlCl_3}+m_{H_2}$

$\Rightarrow m_{HCl}=m_{AlCl_3}+m_{H_2}-m_{Al}$

$\Rightarrow m_{HCl}=13,35+0,15.2-2,7=10,95(g)$

$d,PTHH:2H_2+O_2\xrightarrow{t^o,\text{tia lửa điện}}2H_2O$

$\Rightarrow n_{O_2}=\dfrac{1}{2}n_{H_2}=0,075(mol)$

$\Rightarrow V_{O_2(đktc)}=0,075.22,4=1,68(lít)$

$\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{1,68}{20\%}=8,4(l)$