a.b.Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Fe}=y\end{matrix}\right.\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x 3/2 x ( mol )
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}27x+56y=16,6\\\dfrac{3}{2}x+y=0,5\end{matrix}\right.\) \(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2.27}{16,6}.100=32,53\%\\\%m_{Fe}=100\%-32,53\%=67,47\%\end{matrix}\right.\)
c.\(n_{CuO}=\dfrac{48}{80}=0,6mol\)
\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,6 < 0,5 ( mol )
0,5 0,5 ( mol )
\(m_{Cu}=0,5.64=32g\)
