Đáp án+ Giải thích các bước giải:

`a)`

`PT : Mg + 2HCl -> MgCl_2 + H_2`

         `Fe + 2HCl -> FeCl_2 + H_2`

`b)`

Gọi $\begin{cases} n_{Mg}=a(mol)\\n_{Fe}=b(mol) \end{cases}$ `(a,b>0)`

`=>24a+56b=13,6(1)`

`n_{HCl}={m_{dd}xxC%}/{Mxx100%}={400xx5,475%}/{36,5xx100%}=0,6(mol)`

`=>2a+2b=0,6(2)`

Từ `(1)` và `(2)=>a=0,1;b=0,2`

`=>%m_{Mg}={0,1xx24xx100%}/{13,6}\approx17,65%`

`=>%m_{Zn}=100%-17,65%=82,35%`

`c)`

Theo pt : `n_{H_2}=1/2xx``n_{HCl}=1/2xx``0,6=0,3(mol)`

$m_{\text{dd spứ}}$ `=m_{hh}+m_{HCl}-m_{H_2}`

`=>`$m_{\text{dd spứ}}$`=13,6+400-0,3xx2=413(g)`

`d)`

Dung dịch spứ $\begin{cases} MgCl_2: 0,1(mol)\\FeCl_2:0,2(mol) \end{cases}$

`=>C%`$_{\text{dd $MgCl_2$}}$ `={m_{ct}xx100%}/{m_{dd}}={0,1xx95xx100%}/{413}\approx2,3%`

`=>C%`$_{\text{dd $FeCl_2$}}$ `={m_{ct}xx100%}/{m_{dd}}={0,2xx127xx100%}/{413}\approx6,2%`

$\text{#Cyrus}$

Em tham khảo nha :

Em xem lại đề nha 

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
b)\\
{n_{HCl}} = \dfrac{{400 \times 5,475\% }}{{36,5}} = 0,6mol\\
hh:Mg(a\,mol);Fe(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,6\\
24a + 56b = 13,6
\end{array} \right.\\
 \Rightarrow a = 0,625;b =  - 0,025
\end{array}\)