Используя приведенную ниже диаграмму, определите:

  10,8 г                                                                        Х л

   2Al          +            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

10,8 г Al      -     Х л Н2

54 г Al        -     67,2 л Н2

V(H2) = 10,8 * 67,2 / 54 = 13,44 л

Х г, Х моль                                                                   1 г

  2Al         +           6HCl        =      2 AlCl3         +        3H2

n = 2 моль                                                                  n = 3 моль

Mr = 27 г/моль                                                            Mr = 2 г/моль

m = 54 г                                                                      m = 6 г

Х г Al     -    1 г Н2

54 г Al   -    6 г Н2

m(Al) = 54 * 1 / 6 = 9 г

Х моль Al     -     1 г Н2

2 моль Al     -     6 г Н2

n(Al) = 2 * 1 / 6 = 0,33 моль