\(n_{HCl}=0,65.2=1,3\left(mol\right)\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (1)
0,1<----0,3<------0,1<-----0,15
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\) (2)
x--------->6x-------->2x
\(MgO+2HCl\rightarrow MgCl_2+H_2O\) (3)
y--------->2y-------->y
Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{MgO}=y\left(mol\right)\end{matrix}\right.\)
Theo PTHH (1), (2), (3) có:
\(n_{HCl\left(2\right)}+n_{HCl\left(3\right)}=n_{HCl.ban.đầu}-n_{HCl\left(1\right)}\\ \Leftrightarrow6x+2y=1,3-0,3=1\left(mol\right)\left(I\right)\)
Theo PTHH (1) có: \(m_{Al}=0,1.27=2,7\left(g\right)\\ \Rightarrow m_{Al_2O_3}+m_{MgO}=102x+40y=20,9-2,7=18,2\left(g\right)\left(II\right)\)
Từ (I),(II) có hệ phương trình:
\(\left\{{}\begin{matrix}6x+2y=1\\102x+40y=18,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
Trong hỗn hợp ban đầu:
\(\left\{{}\begin{matrix}m_{Al}=2,7\left(g\right)\\m_{Al_2O_3}=102x=102.0,1=10,2\left(g\right)\\m_{MgO}=40y=40.0,2=8\left(g\right)\end{matrix}\right.\)
b
\(CM_{AlCl_3}=\dfrac{0,1+2x}{0,65}=\dfrac{0,1+2.0,1}{0,65}=\dfrac{6}{13}\approx0,46M\)
\(CM_{MgCl_2}=\dfrac{y}{0,65}=\dfrac{0,2}{0,65}=\dfrac{4}{13}\approx0,31M\)