\(2Al+3H_2SO_4 \to Al_2(SO_4)_3+3H_2\\ n_{Al}=0,4(mol)\\ a/\\ n_{H_2}=\frac{3}{2}.0,4=0,6(mol)\\ V_{H_2}=0,6.22,4=13,44(l)\\ b/\\ n_{H_2SO_4}=n_{H_2}=0,6(mol)\\ n_{ddH_2SO_4}=\frac{0,6.98.100}{20}=294(g)\\ c/\\ n_{Al_2(SO_4)_3}=0,2(mol)\\ C\%_{Al_2(SO_4)_3}=\frac{0,2.342}{10,8+294-0,6.2}.100\%=22,52\%\)