`n_(Al)=\frac{2,7}{27}=0,1(mol)`
`a,`
`2Al+6HCl->2AlCl_3+3H_2`
`0,1` `0,15` `0,1` `0,15`
`b`
`m_(AlCl_3)=0,1.133,5=13,35(g)`
`c,`
`V_(H_2)=0,15.22,4=3,36(l)`
`m_(H_2)=0,15.2=0,3(mol)`
`d,`
`m_(HCl)=0,15.36,5=5,475(g)`
`=>m_(dd HCl)=\frac{5,475}{3,65%}=150(g)`
`m_(dd)=2,7+150-0,3=152,4(g)`