Đáp án:

`↓`

Giải thích các bước giải:

`a,` `PTHH:`

          `2Al``+``6HCl``→``2AlCl_3``+``3H_2↑`

`b,` `m_{HCl}=``\frac{C%.mdd}{100%}``=``\frac{10%.195}{100%}``=``19,5` `(g)`

     `n_{HCl}=``\frac{m}{M}``=``\frac{19,5}{36,5}``≈``0,53` `(mol)`

`PTHH`            `2Al``+``6HCl``→``2AlCl_3``+``3H_2↑` 

                      `0,18`     `0,53`        `0,18`        `0,265`        `mol`

`→n_{Al}=``\frac{2}{6}n_{HCl}``=``\frac{0,53.2}{6}``≈``0,18` `(mol)`

`→m_{Al}=n.M=0,18.27=4,86` `(g)`

`c,`Theo pt, ta có: `n_{H_2}=``\frac{3}{6}n_{HCl}``=``\frac{0,53.3}{6}``=``0,265` `(mol)`

`→V_{H_2(đktc)}=n.22,4=0,265.22,4=5,936` `(l)`

`d,` `m_{H_2}=n.M=0,265.2=0,53` `(g)`

`m_{sau}=m_{Al}+m_{dd_{HCl}}-m_{H_2}=4,86+195-0,53=199,33` `(g)`

Theo pt, ta có: `n_{AlCl_3}=``\frac{2}{6}n_{HCl}``=``\frac{0,53.2}{6}``≈``0,18` `(mol)`

`→m_{AlCl_3}=n.M=0,18.133,5=24,03` `(g)`

`⇒C%_{AlCl_3}=``\frac{m.100%}{m_{sau}}``=``\frac{24,03.100%}{199,33}``≈``12,06%`

2Al + 2H₂SO₄ → Al₂(SO₄)₃ + 3H₂

 

Giải thích các bước giải: