\(PTHH:MgO+H_2SO_4\rightarrow MgSO_4+H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.\dfrac{100.24\%}{40}=0,3\left(mol\right)\\ \Rightarrow n_{H_2SO_4pứ}=1-0,3=0,7\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{MgO}=x\left(mol\right)\\n_{Fe_2O_3}=y\left(mol\right)\end{matrix}\right.\\ Tacó:40x+160y=32\\ Mặckhác:x+3y=0,7\\ \Rightarrow x=0,4;y=0,1\\ \Rightarrow\%MgO=\dfrac{0,4.40}{32}.100=50\%;\%Fe_2O_3=50\%\)
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