Fe+2HCl--->FeCl2 +H2
x-----------------x
MgO +2HCl----->MgCl2 +H2
y------------------------y
Ta có
\(\left\{{}\begin{matrix}56x+40y=13,6\\127x+95y=31,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
%m\(_{Fe}=\frac{0,1.56}{13,6}.100\%=41,18\%\)
%m\(_{MgO}=100-41,18=58,52\left(g\right)\)
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