PTHH: 2Al + 3S ---t--> Al2S3
          2Al + 6HCl ----> 2AlCl3 + 3H2
           Al2S3 + 3HCl -----> AlCl3 + 3H2S
nS = 0,96/32=0,03 mol
nAl = 1,08/27=0,04 mol
Vì 0,04/2 > 0,03/3 
=> Sau pứ, Al dư, S hết
nAl pứ=0,03.2/3=0,02 mol
=> nAl dư = 0,04 - 0,02 = 0,02 mol
=> nH2= 0,02.3/2=0,03 mol => mH2=0,03.2=0,06 g
nAl2S3=0,03/3=0,01 mol
=> nH2S=0,01.3=0,03 mol => mH2S=0,03.34=1,02g
=> mX=mH2+mH2S=0,06+1,02=1,08g
%mH2=0,06/1,08.100%=5,56%
%mH2S=100% - 5,56%=94,44%