Đáp án:

`↓`

Giải thích các bước giải:

`a)` `n_{Al}``=``\frac{m}{M}``=``\frac{2,7}{27}``=``0,1` `(mol)`

`PTHH`           `2Al``+``3H_2SO_4``→``Al_2(SO_4)_3``+``3H_2`

                       `0,1`                                `0,05`             `0,15`        `mol`

`→``n_{Al_2(SO_4)_3}``=``n_{Al}``=``\frac{0,1.1}{2}``=``0,05` `(mol)`

     `m_{Al_2(SO_4)_3}``=``n``.``M``=``0,05``.``342``=``17,1` `(g)`

`→``n_{H_2}``=``n_{Al}``=``\frac{0,1.3}{2}``=``0,15` `(mol)`

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

`b)` `V_{sau}``=``V_{H_2SO_4}``=``300` `(ml)``=``0,3` `(l)`

`→`$C_{M_{Al_2(SO_4)_3}}$`=``\frac{n}{V_{sau}}``=``\frac{0,05}{0,3}``≈``0,167` `(M)`

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

$PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\uparrow$

Theo PT: $n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,05(mol)$

$\Rightarrow m_{Al_2(SO_4)_3}=0,05.342=17,1(g)$

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

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

$b)$ Đổi $300ml=0,3l$

$\Rightarrow C_{M_{Al_2(SO_4)_3}}=\dfrac{0,05}{0,3}\approx 0,17M$