\(Fe+2HCl\rightarrow FeCl_2+H_2\)
x____2x__________________
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
y_____3y___________________
Đổi : 600ml=0,6l
\(n_{HCl}=0,6.1,5=0,9\left(mol\right)\)
Giải hệ phương trình :
\(\left\{{}\begin{matrix}56x+27y=33\\2x+3y=0,9\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,6\\y=-0,13\end{matrix}\right.\)
\(\rightarrow\) X không tan hết du có nghiêm âm
\(n_{HCl_{moi}}=1,6.1,5=2,4\)
\(\left\{{}\begin{matrix}56x+27y=33\\2x+3y=2,4\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,6\end{matrix}\right.\)
\(m_{Fe}=0,3.56=16,8\left(g\right)\)
\(m_{Al}=0,6.27=16,2\left(g\right)\)
