a.Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Fe}=y\end{matrix}\right.\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(Mg+H_2SO_4\left(l\right)\rightarrow MgSO_4+H_2\)
x x ( mol )
\(Fe+H_2SO_4\left(l\right)\rightarrow FeSO_4+H_2\)
y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}24x+56y=6,8\\x+y=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,05.24}{6,8}.100=17,64\%\\\%m_{Fe}=\dfrac{0,1.56}{6,8}.100=82,36\%\end{matrix}\right.\)
b.\(Mg+2H_2SO_4\left(đ\right)\rightarrow MgSO_4+SO_2+2H_2O\)
0,05 0,05 ( mol )
\(V_{SO_2}=0,05.22,4=1,12l\)
