2Al+6HCl->2AlCl3+3H2
0,2-----0,6------0,2-----0,3 mol
nAl=\(\dfrac{5,4}{27}\)=0,2 mol
=>VH2=0,3.22,4=6,72l
=>m HCl=0,6.36,5=21,9g
=>m AlCl3=0,2.133,5=26,7g
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ 2Al+6HCl\rightarrow\left(t^o\right)2AlCl_3+3H_2\\ n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\\ n_{HCl}=\dfrac{6}{2}.0,2=0,6\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{HCl}=0,6.36,5=21,9\left(g\right)\\ c,m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\)