Calculate the maximum mass in grams of h2s that can form when 165 g aluminum sulfide reacts with 125 g water according to:

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