`a)`

$\ Mg+2HCl\to MgCl_2+H_2\\\hspace{0,1cm}0,3\hspace{0,1cm}\leftarrow\hspace{0,1cm}0,6\hspace{0,1cm}\leftarrow\hspace{0,3cm}0,3\leftarrow\hspace{0,3cm}0,3(mol)$ $\ \ MgO+2HCl\to MgCl_2+H_2O\\\hspace{0,1cm}0,075\hspace{0,1cm}(0,75-0,6)\hspace{0,1cm}0,075\hspace{1,5cm}(mol)$

`500(ml)=0,5(l)`

`n_{HCl}=0,5.1,5=0,75(mol)`

`n_{H_2}={6,72}/{22,4}=0,3(mol)`

`m_{Mg}=0,3.24=7,2(g)`

`m_{MgO}=0,075.40=3(g)`

`m_X=7,2+3=10,2(g)`

`b)`

`V_{dd\ spu}=V_{dd\ HCl}=0,5(l)`

`n_{MgCl_2}=0,3+0,075=0,375(mol)`

`C_{M\ MgCl_2}={0,375}/{0,5}=0,75(M)`

`n_(H_2` `=(2,24)/(22,4)=0,1(mol)`

`400(ml)=0,4(l)`

`n_(HCl` `=0,4.2=0,8(mol)`

Do thu được kim loại

`->` Muối tạo thành chỉ có `FeCl_2`

   `Fe+2HCl->FeCl_2+H_2` $(1)$

   `Fe_2O_3+6HCl->2FeCl_3+3H_2O` $(2)$

   `Fe+2FeCl_3->3FeCl_2` $(3)$

Theo pt $(1):$

`n_{Fe}=n_{H_2}=0,1(mol)`

`n_{HCl}=2n_{H_2}=0,2(mol)`

`n_(FeCl_2` `=n_(H_2` `=0,1(mol)`

Theo pt $(2):$ 

`n_{Fe_2O_3}=(1)/(6)n_{HCl}=(1)/(6).(0,8-0,2)=0,1(mol)`

`n_(FeCl_3` `=2n_{Fe_2O_3}` `=0,2(mol)`

Theo pt $(3):$

`n_{Fe}=(1)/(2)n_{FeCl_3}=0,1(mol)`

`n_{FeCl_2}=3n_{Fe}=0,3(mol)`

`->m_(\text{muối(Y)}` `=127.(0,1+0,3)=50,8(g)`

`m_{X}=m_{Fe(1)}+m_{Fe_2O_3}+m_{Fe(3)}+m_{\text{KL thu được}}`

`=0,1.56+0,1.160+0,1.56+2,8=30(g)`