\(PTHH:\)
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
\(Mg+2H_2SO_4\rightarrow MgSO_4+SO_2+2H_2O\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}:x\left(mol\right)\\n_{Mg}:y\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow56x+24y=10,4\left(g\right)\)
Ta có:
\(n_{SO2}=\frac{3}{2}n_{Fe}+n_{Mg}=1,5x+y=\frac{7,84}{22,4}=0,35\left(mol\right)\)
\(\Rightarrow n_{Fe2\left(SO4\right)3}=\frac{1}{2}n_{Fe}=0,05\left(mol\right)\)
\(\Rightarrow n_{MgSO4}=n_{Mg}=0,2\left(mol\right)\Rightarrow m_{hh}=120\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Mg}=0,2.24=4,8\left(g\right)\end{matrix}\right.\)