\(Al_2O_3+3H_2SO_3\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
Ta có: \(n_{Al2O4}=10,2.\left(27.2+16.3\right)=0,1\left(mol\right)\)
\(n_{Mg}=\frac{4}{24+16}=0,1\left(mol\right)\)
Theo phản ứng:n H2SO4=3nAl2O3 + nMgO\(=0,1.2+0,1=0,4\left(mol\right)\)
\(\rightarrow V_{H2SO4}=\frac{0,4}{2}=0,2\left(l\right)\)