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If 48.1 g of solid Al2S3 and 6.68 mL of liquid H2O are reacted stoichiometrically according to the balanced equation, how many grams of solid Al2S3 remain?

Al2S3(s) + 6H2O(l) → 2Al(OH)3(s) + 3H2S(g)

I found that the limiting reagent,
.6406mol of Al2S3 and .1226mol of H2O
what do I do next

  • Chemistry -

    You have 48.1/150.161 = 0.3203 mols Al2S3.
    You have 6.00g/18.015 = 0.333 moles H2O.

    The problem tells you H2O is the limiting reagent so convert moles H2O to moles Al2S3 to see how much will be used.
    0.333 x (1 mole Al2S3/6 moles H2O) = 0.333 x (1/6) = 0.0555 moles.

    Convert 0.0555 moles Al2S3 used to grams (g = moles x molar mass) and subtract from the 48.1 g to find the amount Al2S3 unreacted.

  • Chemistry -

    ...I got 39.76..I got it wrong..

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