Mg+H2SO4->MgSO4+H2
x------------------------------x
Fe+H2SO4->FeSO4+h2
y---------------------------------y
nH2=6,72\22,4=0,3 mol
=>24x+56y=13,6
......x+y=0,3
=>x=0,1 , y=0,2 mol
=>%mMg=0,1.24\13.6.100=17,6%
=>%mFe=100-17,6=82,4%
Ta có:
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:a\left(mol\right)\\n_{Mg}:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}56a+24b=13,6\\a+b=0,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{11,2}{13,6}.100\%=82,35\%\\\%m_{Mg}=100\%-82,35\%=17,65\%\end{matrix}\right.\)
