\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ a,Mg+2H_2SO_4\to MgSO_4+SO_2\uparrow+2H_2O(1)\\ MgO+H_2SO_4\to MgSO_4+H_2O(2)\\ \Rightarrow n_{Mg}=n_{MgSO_4(1)}=0,4(mol)\\ \Rightarrow m_{Mg}=0,4.24=9,6(g)\\ n_{MgSO_4(2)}=\dfrac{72-0,4.120}{120}=0,2(mol)\\ \Rightarrow n_{MgO}=0,2(mol)\\ \Rightarrow m_{hh}=0,2.56+9,6=20,8(g)\\ b,\Sigma n_{H_2SO_4}=2.0,4+0,2=1(mol)\\ \Rightarrow C\%_{H_2SO_4}=\dfrac{1.98}{122,5}.100\%=80\%\)
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