+\(n_{Al_2O_3}=\frac{20,4}{102}=0,2\left(mol\right)\)
PTHH: \(Al_2O_3+3H_2SO_4\)→\(Al_2\left(SO_4\right)_3\)+\(3H_2O\)
0,2_______0,3______0,2_______0,3(mol)
+\(m_{H_2SO_4}=0,2.98=19,6\left(gam\right)\)
+\(m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(gam\right)\)
+\(V_{H_2}=0,2.22,4=4,48\left(mol\right)\)
:))