3,65 г                                      Х л

  2Al        +         6HCl      =       2AlCl3      +      3H2

                       n = 6 моль                               n = 3 моль

                    Mr = 36 г/моль                           Vm=22,4 л/моль

                    m = 212 г                                   V = 67,2 л

3,65 г HCl    -      Х л Н2

212 г HCl    -      67,2 л Н2

V(H2) = 3,65 * 67,2 / 212 = 1,15 л