`a)`
`2Al+6HCl →2AlCl_3 +3H_2`
`b)`
`n_(H_2)= (1,12)/(22,4)=0,05(mol)`
Theo PTHH
`n_(H_2)=2n_(HCl)=0,05.2=0,1(mol)`
`⇒m_(HCl)=``0,1.36,5``=``3,65(g)`
Theo PTHH
`n_(H_2)=3/2n_(Al)=3/2.0,05=1/30(mol)`
`⇒m_(Al)=1/30.27=0,9(g)`
`c)`
Theo PTHH
`n_(Al)=n_(AlCl_3)=1/30(mol)`
`⇒m_(AlCl_3)=1/30. 133,5=4,45(g)`