Respuesta :

Al₂S₃ + 6H₂O = 2Al(OH)₃ + 3H₂S

n(Al₂S₃)=m(Al₂S₃)/M(Al₂S₃)
n(Al₂S₃)=165g/150.16g/mol=1.10 mol

n(H₂O)=m(H₂O)/M(H₂O)
n(H₂O)=125g/18.02g/mol=6.94 mol

1.10:6.94 = 1:6.31
The limiting reagent is an aluminum sulfide.

m(H₂S)=n(H₂S)M(H₂S)=3n(Al₂S₃)M(H₂S)
m(H₂S)=3*1.10mol*34.08g/mol=112.46 g

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