\(n_{H_2}=\dfrac{11,76}{22,4}=0,525\left(mol\right)\)
Đặt \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Al}=y\end{matrix}\right.\) ( mol )
\(\Rightarrow m_{hh}=56x+27y=15,15\left(1\right)\)
`Fe+H_2SO_4->FeSO_4+H_2`
x x ( mol )
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H2`
y 1,5y ( mol )
\(n_{H_2}=x+1,5y=0,525\left(2\right)\)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,25\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,15.56}{15,15}.100\%=55,44\%\\\%m_{Al}=100\%-55,44\%=44,56\%\end{matrix}\right.\)
\(15,5g\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\xrightarrow[9,8\%]{+H_2SO_4}\left\{{}\begin{matrix}Al_2\left(SO_4\right)_3:\dfrac{x}{2}\left(mol\right)\\FeSO_4:y\left(mol\right)\end{matrix}\right.+H_2:0,525\left(mol\right)\)
\(BTKL:27x+56y=15,15\)
Xét quá trình nhường nhận electron:
\(\begin{matrix}Fe^0\rightarrow Fe^{2+}+2e\\Al^0\rightarrow Al^{3+}+3e\end{matrix}\text{ | }2H^++2e\rightarrow H_2\)
\(BT\left(e\right):3x+2y=0,525\)
Lập hệ phương trình: \(\left\{{}\begin{matrix}27x+56y=15,15\\3x+2y=0,525\cdot2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,25\\y=0,15\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,25\cdot27=6,75\left(g\right)\\m_{Fe}=0,15\cdot56=8,4\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{6,75}{15,15}\cdot100\%\approx44,5\%\\\%m_{Fe}\approx100\%-44,5\%=55,5\%\end{matrix}\right.\)