\(n_{Mg}=x;n_{Fe}=y\\ PTHH:Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ hpt:\left\{{}\begin{matrix}24x+56y=8\\22,4\left(x+y\right)=4,48\end{matrix}\right.\Leftrightarrow x=y=0,1\\ \rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\frac{0,1.24}{8}.100\%=30\%\\\%m_{Fe}=1100-30=70\%\end{matrix}\right.\\ n_{axit}=x+y+15\%.\left(x+y\right)=0,1+0,1+15\%.\left(0,1+0,1\right)=0,23\left(mol\right)\\ \rightarrow V_{axit}=\frac{0,23}{0,5}=0,46\left(l\right)\)
Mg + H2SO4----> MgSO4 +H2
x--------x----------------x-------x
Fe+H2SO4----> FeSO4 +H2
y-----y------------y---------y
a) Gọi n\(_{Mg}=x\Rightarrow m_{Mg}=24x\)
n\(_{Fe}=y\Rightarrow m_{Fe}=56y\)
=> 24x +56y=8(*)
Mặt khác n\(_{H2}=\frac{4,48}{22,4}=0,2mol\)
=> x+y=0,1( **)
Từ (1) và (2) ta có hệ pt
\(\left\{{}\begin{matrix}24x+56y=8\\x+y=0,1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
%m\(_{Mg}=\frac{0,1.2,4}{8},100\%=\)30%
%m\(_{Fe}=100\&-30\%=70\%\)
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