\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{9,6}{24}=0,4mol\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\)
\(2Al+6HCl\rightarrow AlCl_3+3H_2\)
x 1/2x 3/2x ( mol )
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}27x+56y=11\\\dfrac{3}{2}x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Al}=0,2.27=5,4g\)
\(\Rightarrow m_{Fe}=0,1.56=5,6g\)
\(\%m_{Al}=\dfrac{5,4}{11}.100=49,09\%\)
\(\%m_{Fe}=100\%-49,09\%=50,91\%\)
\(m_{muối.thu.được}=\left(\dfrac{1}{2}.0,2\right).133,5+0,1.127=26,05g\)
