\(1)\\ a)n_{H_2SO_4}=\dfrac{98}{98}=1\left(mol\right)\\ Mg+H_2SO_4\xrightarrow[]{}MgSO_4+H_2\\ \Rightarrow n_{H_2SO_4}=n_{H_2}=1mol\\ V_{H_2}=1.22,4=22,4\left(l\right)\\ b)m_{ddH_2SO_4}=\dfrac{98}{20\%}\cdot100\%=490\left(g\right)\)
\(a)n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\\ m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\ b)n_{HCl}=\dfrac{0,2.6}{2}=0,6\left(mol\right)\\ m_{HCl}=0,6.36,5=21,9\left(g\right)\\ m_{ddHCl}=\dfrac{21,9}{30\%}.100\%=73\left(g\right)\\ n_{H_2}=\dfrac{0,2.3}{2}=0,3\left(mol\right)\\ m_{H_2}=0,3.2=0,6\left(g\right)\\ m_{ddAlCl_3}=5,4+73-0,6=77,8\left(g\right)\\ C\%_{AlCl_3}=\dfrac{26,7}{77,8}.100\%\approx34\%\)
