$a,n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{H_2}=n_{Fe}=0,1(mol)\\ \Rightarrow V_{H_2}=0,1.22,4=2,24(l)\\ c,n_{HCl}=2n_{Fe}=0,2(mol);n_{FeCl_2}=n_{Fe}=0,1(mol)\\ C_1:m_{FeCl_2}=0,1.127=12,7(g)\\ C_2:\text{Bảo toàn KL: }\\ m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow 5,6+0,2.36,5=m_{FeCl_2}+0,1.2\\ \Rightarrow m_{FeCl_2}+0,2=12,9\\ \Rightarrow m_{FeCl_2}=12,7(g)\\ c,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}(mol)\\ \Rightarrow m_{Al}=\dfrac{1}{15}.27=1,8(g)$

`a.` `nFe``=``\frac{m}{M}``=``\frac{5,6}{56}``=``0,1` `(mol)`

`PTHH`          `Fe``+``2HCl``→``FeCl_2``+``H_2`

                        `0,1`     `0,2`         `0,1`         `0,1`    `mol`

`→``nH_2``=``nFe``=``0,1` `(mol)`

`→``VH_2``(đktc)``=``n``.``22,4``=``0,1``.``22,4``=``2,24` `(l)` 

`b.`

Cách 1:

Theo pt, ta có: `nFeCl_2``=``nFe``=``0,1` `(mol)`

`→``mFeCl_2``=``n``.``M``=``0,1``.``127``=``12,7` `(g)`

Cách 2:

`mH_2``=``n``.``M``=``0,1``.``2``=``0,2` `(g)`

Theo pt, ta có: `nHCl``=``nFe``=``\frac{0,1.2}{1}``=``0,2` `(mol)`

`→``mHCl``=``n``.``M``=``0,2``.``36,5``=``7,3` `(g)`

Áp dụng ĐLBTKL, ta có:

        `mFe``+``mHCl``=``mFeCl_2``+``mH_2`

  `⇔``5,6``+``7,3``=``mFeCl_2``+``0,2`

  `⇔``12,9``=``mFeCl_2``+``0,2`

  `⇔``mFeCl_2``=``12,7` `(g)`

`c.` 

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

                 `0,067`                                              `0,1`      `mol`

`→``nAl``=``nH_2``=``\frac{0,1.2}{3}``≈``0,067` `(mol)`

`→``mAl``=``n``.``M``=``0,067``.``27``=``1,809` `(g)`