Ta có: \(n_{Al_2O_3}=\dfrac{120}{102}=\dfrac{20}{17}\left(mol\right)\)
PT: \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=3n_{Al_2O_3}=\dfrac{60}{17}\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=n_{Al_2O_3}=\dfrac{20}{7}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{\dfrac{60}{17}}{0,3}=\dfrac{200}{17}\left(M\right)\)
\(m_{Al_2\left(SO_4\right)_3}=\dfrac{20}{7}.342=\dfrac{6840}{7}\left(g\right)\)