Verified answer

  5,4 г                                              Х г

   2Al         +         3S          =          Al2S3

n = 2 моль                                   n = 1 моль

Mr = 27 г/моль                            Mr = 150 г/моль

m = 54 г                                       m = 150 г

5,4 г Al    -     Х г Al2S3

54 г Al    -     150 г Al2S3

m(Al2S3) = 5,4 * 150 \ 54 = 15 г