Đáp án + Giải thích các bước giải:

$a\big)$

Đặt $n_{Mg}=x(mol);n_{Al}=y(mol)$

$\to 24x+27y=9,6(1)$

$Mg+2HCl\to MgCl_2+H_2$

$2Al+6HCl\to 2AlCl_3+3H_2$

Theo PT: $x+1,5y=n_{H_2}=\frac{11,2}{22,4}=0,5(2)$

Từ $(1)(2)\to x=0,1(mol);y=\frac{4}{15}(mol)$

$\to \%m_{Mg}=\frac{0,1.24}{9,6}.100\%=25\%$

$\to \%m_{Al}=100-25=75\%$

$b\big)$

Theo PT: $n_{HCl}=2n_{H_2}=1(mol)$

$\to C\%_{HCl}=\frac{1.36,5}{146}.100\%=25\%$

$c\big)$

Theo PT: $n_{MgCl_2}=x=0,1(mol);n_{AlCl_3}=y=\frac{4}{15}(mol)$

$\to m_{MgCl_2}=0,1.95=9,5(g);m_{AlCl_3}=\frac{4}{15}.133,5=35,6(g)$

$\to m_{muối}=9,5+35,6=45,1(g)$

Đáp án+ Giải thích các bước giải:

$a)$

$Mg + 2HCl$ $\rightarrow$ $MgCl_2 + H_2(1)$
$2Al + 6HCl$ $\rightarrow$ $2AlCl_3 + 3H_2(2)$

Gọi $a,b$ lần lượt là $n_{Mg}$ , $n_{Al}$ $(a,b>0)$

 Ta có: $\begin{cases} 24a+27b=9,6\\a+1,5b= \dfrac{11,2}{22,4} \end{cases}$ $\rightarrow$ $\begin{cases} a=0,1\\b=\dfrac{4}{15} \end{cases}$ 

$\rightarrow$ $\text{%$m_{Mg}$= $\frac{0,1.24.100}{9,6}$= 25% }$ 

$\rightarrow$ $\text{%$m_{Al}$= 100-25=75%}$ 

$b)$

Theo pt $(1)$ và $(2)$: $\Sigma$$n_{HCl}$= $0,1.2+ \dfrac{4}{15}.3= 1(mol)$

$\rightarrow$ $\text{C%}_{dd HCl}$ = $\frac{1.36,5.100}{146}$ = $\text{25%}$

$c)$

Theo pt $(1)$ : $n_{MgCl_2}$= $n_{Mg}$= $0,1(mol)$

Theo pt $(2)$ : $n_{AlCl_3}$= $n_{Al}$= $\dfrac{4}{15}(mol)$

$\rightarrow$ $m_{muối}$= $m_{MgCl_2}$ + $m_{AlCl_3}$ = $0,1.95+ \dfrac{4}{15}.133,5=45,1(g)$