$PTHH:Mg+2HCl→MgCl_2+H_2$ $(1)$

$2Al+6HCl→2AlCl_3+3H_2$ $(2)$

a. Gọi $n_{Mg}=a(mol),n_{Al}=b(mol)$

$m_{hh}=7,8g→24a+27b=7,8$ $(*)$

Theo $PTHH:n_{H_2\ (1)}=a(mol)$

$n_{H_2\ (2)}=1,5b(mol)$

Tổng $n_{H_2}=8,96:22,4=0,4(mol)$

$→a+1,5b=0,4$ $(**)$

Từ $(*)$ và $(**)→\left \{ {{a=0,1\ (mol)} \atop {b=0,2\ (mol)}} \right.$ 

$→m_{Mg}=0,1.24=2,4g$

$→\%Mg=\frac{2,4}{7,8}.100\%=30,77\%$

$\%Al=100-30,77=69,23\%$

b. Theo $PTHH:n_{MgCl_2}=n_{Mg}=0,1(mol)$

$n_{AlCl_3}=n_{Al}=0,2(mol)$

$→m_{muối}=0,1.95+0,2.133,5=36,2g$

Đáp án:

\(a,\ \%m_{Mg}=30,77\%\\ \%m_{Al}=69,23\%\\ b,\ m_{\text{muối khan}}=36,2\ g.\)

Giải thích các bước giải:

\(10/\\ a,\ PTHH:\\ Mg+2HCl\to MgCl_2+H_2↑\ (1)\\ 2Al+6HCl\to 2AlCl_3+3H_2↑\ (2)\\ n_{H_2}=\dfrac{8,96}{22,4}=0,4\ mol.\\ \text{Gọi $n_{Mg}$ là a (mol), $n_{Al}$ là b (mol)}\\ \text{Theo đề bài ta có hệ pt:}\\ \left\{\begin{matrix} 24a+27b=7,8 & \\ a+1,5b=0,4 & \end{matrix}\right.\Rightarrow \left\{\begin{matrix} a=0,1 & \\ b=0,2 & \end{matrix}\right.\\ \Rightarrow \%m_{Mg}=\dfrac{0,1.24}{7,8}\times 100\%=30,77\%\\ \%m_{Al}=\dfrac{0,2.27}{7,8}\times 100\%=69,23\%\\ b,\ Theo\ pt\ (1):\ n_{MgCl_2}=n_{Mg}=0,1\ mol.\\ Theo\ pt\ (2):\ n_{AlCl_3}=n_{Al}=0,2\ mol.\\ \Rightarrow m_{MgCl_2}=0,1.95=9,5\ g.\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7\ g.\\ \Rightarrow m_{\text{muối khan}}=m_{MgCl_2}+m_{AlCl_3}=9,5+26,7=36,2\ g.\)

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