%mMgO=0.3⋅4028⋅100%=42.85%%mMgO=0.3⋅4028⋅100%=42.85%
%mFe2O3=100−42.85=57.15%
nMgO=a(mol),nFe2O3=b(mol)nMgO=a(mol),nFe2O3=b(mol)
⇒mhh=40a+160b=28(g)(1)⇒mhh=40a+160b=28(g)(1)
\(n_{H_2SO_4}=\dfrac{200.31,5\%}{98}=\dfrac{9}{14}\left(mol\right)\\ \left\{{}\begin{matrix}n_{ZnO}=a\left(mol\right)\\n_{Al_2O_3}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ ZnO+H_2SO_4\rightarrow ZnSO_4+H_2\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ \rightarrow\left\{{}\begin{matrix}81a+102b=28,5\\a+3b=\dfrac{9}{14}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{93}{658}\\b=\dfrac{55}{329}\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{ZnO}=\dfrac{\dfrac{93}{658}.81}{28,5}.100\approx\\\%m_{Al_2O_3}\approx59,83\%\end{matrix}\right.40,17\%}\)
