1,5 г                                                                             Х л

   Al2S3        +         6H2O      =       2Al(OH)3        +     3 H2S

n = 1 моль                                                                 n = 3 моль

Mr = 150 г/моль                                                       Vm = 22,4 л/моль

m = 150 г                                                                  V = 67,2 л

1,5 г Al2S3     -     Х л H2S

150 г Al2S3    -    67,2 л H2S

V(H2S) = 1,5 * 67,2 / 150 = 0,672 л

Ответ: а