\(Mg+2HCl-->MgCl2+H2\)
x---------2x-----------------------------x(mol)
\(2Al+6HCl-->2AlCl3+3H2\)
y-----------3y------------------------1,5y(mol)
Ta có: \(m_{hh}=7,8\left(g\right)\Rightarrow24x+27y=7,8\)
\(\Rightarrow9\left(2x+3y\right)+6x=7,8\)
\(\Rightarrow\left(2x+3y\right)+6x=0,8667\)
\(\Rightarrow2x+3y=0,8667-6x\)
Do \(x\ge0\Rightarrow2x+3y\le0,8667\) ( 1)
Mặt khác
\(m_{HCl}=\frac{200.18,25}{100}=36,5\left(g\right)\)
\(n_{HCl}=\frac{36,5}{36,5}=1\left(mol\right)\)
hay \(2x+3y=1\) (2)
Từ 1 và 2 \(\Rightarrow HCl\) dư...KL hết ( mục đích tìm chất hết để giải hệ và tính theo chất hết nha)
\(n_{H2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+27y=7,8\\x+1,5y=0.4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%m_{Mg}=\frac{0,1.24}{7,8}.100\%=30,77\%\)
\(\%m_{Al}=100-30,77=69,23\%\)
b) \(m_{dd}\) sau pư = \(m_{KL}+m_{ddHCl}-m_{H2}=7,8+200-0,8=207\left(g\right)\)
\(n_{MgCl2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgCl2}=0,1.95=9,5\left(g\right)\)
\(C\%_{MgCl2}=\frac{9,5}{207}.100\%=4,59\%\)
\(n_{Alcl3}=n_{Al}=0,2\left(mol\right)\)
\(m_{AlCl3}=0,2.133,5=26,7\left(g\right)\)
\(C\%_{AlCl3}=\frac{26,7}{207}.100\%=12,9\%\)