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29 December, 15:25

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

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  1. 29 December, 15:47
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    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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