18,6 г                                                            Х л

 2 Al         +       6HCl   =     2AlCl3      +      3H2

n = 2 моль                                                n = 3 моль

Mr = 27 г/моль                                        Vm = 22,4 л/моль

m = 54 г                                                  V = 67,2 л

18,6 л Al      -     Х л Н2

54 г Al         -    67,2 л Н2

V(H2) = 18,6 * 67,2 / 54 =23,15 л