Răspuns:
M,Al= 24g/mol
M,S=32g/mol
M,Al2S3= 150g/mol
2Al + 3S=====> Al2S3
m,Al/m,S= 54/96
m,Al/m,S= 5,4/6,4,
se observa ca S este intr-o cantitate mai mica, iar Al este in exces
54gAl..........96gS.........150gAl2S3
X..................6,4g................y
x= 3,6gAl intra in reactie, si ramane, m= 5,4-3,6g nereactionat
y=10gAl2S3
150g...........6mol
Al2S3 + 6HCl======> 2AlCl3 + 3H2S
10g.............x= 0,4mol HCl
c=niu/V---> V= niu/c= 0,4mol/2mol/l= 0,2 l= 200ml solutieHCl