a,
Hỗn hợp: $\begin{cases}\rm Al:x\ (mol)\\\rm Zn:y\ (mol)\end{cases}$
$\rm\to 27x+65y=4\ \ (1)$
$\rm n_{H_2}=\dfrac{3,808}{22,4}=0,17\ \ (mol)$
Bán phản ứng:
$\rm \mathop{Al}\limits^{0}-3e\to \mathop{Al}\limits^{+3}\qquad ||\qquad \mathop{2H}\limits^{+1}+2e\to \mathop{H_2}\limits^{0}\\\ x\ \ \ \ \ \ 3x \qquad\qquad\qquad\qquad\quad0,34 \ \ \ 0,17\\\mathop{Zn}\limits^{0}-2e\to\mathop{Zn}\limits^{+2}\\ \ y \ \ \ \ \ \ 2y$
Áp dụng định luật bảo toàn e:
$\rm e_{nhường}=e_{nhận}$
$\rm\to 3x+2y=0,34 \ \ (2)$
Từ `(1)` và `(2)` ta có hệ: $\begin{cases}\rm27x+65y=4\\\rm3x+2y=0,34\end{cases}$
`<=>` $\begin{cases}\rm x=0,1\ (mol)\\\rm y=0,02 \ (mol)\end{cases}$ `<=>` $\begin{cases}\rm m_{Al}=0,1.27=2,7\ (g)\\\rm m_{Zn}=0,02.65=1,3\ (g)\end{cases}$
$\rm \%m_{Al}=\dfrac{2,7}{4}.100\%=67,5\%$
$\rm \%m_{Zn}=100\%-67,5\%=32,5\%$
b, Ta có:
$\rm n_{AlCl_3}=n_{\mathop{Al}\limits^{+3}}=0,1\ (mol)$
$\rm\to m_{AlCl_3}=0,1.133,5=13,35\ (g)$
$\rm n_{ZnCl_2}=n_{\mathop{Zn}\limits^{+2}}=0,02\ (mol)$
$\rm\to m_{ZnCl_2}=0,02.136=2,72\ (g)$
$\rm m_{muối}=13,35+2,72=16,07\ (g)$
c, Lại có:
$\rm n_{HCl}=3n_{AlCl_3}+2n_{ZnCl_2}=3.0,1+2.0,02=0,34\ (mol)$
$\rm V_{HCl}=\dfrac{0,34}{0,2}=1,7\ (M)$
