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

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.

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

To find out how many grams of solid Al2S3 remain after the reaction, you first need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.

In this case, you've already calculated that there are 0.6406 mol of Al2S3 and 0.1226 mol of H2O. To determine the limiting reagent, you need to compare the molar ratios of the reactants in the balanced equation.

From the balanced equation: Al2S3(s) + 6H2O(l) → 2Al(OH)3(s) + 3H2S(g),
the molar ratio of Al2S3 to H2O is 1:6.

Since the amount of H2O is much less compared to the amount required for the reaction (1:6 ratio), it is the limiting reagent.

Now, to find the amount of Al2S3 that reacts, we can use stoichiometry. From the balanced equation, we see that 1 mol of Al2S3 reacts with 6 mol of H2O.

So, if 0.1226 mol of H2O is used, we can calculate the moles of Al2S3 by taking (0.1226 mol of H2O) * (1 mol of Al2S3 / 6 mol of H2O) = 0.0204 mol of Al2S3.

Now, to find the mass of Al2S3 remaining, we can subtract the moles of Al2S3 reacted from the initial moles of Al2S3.
Initial moles of Al2S3 = 0.6406 mol, and moles of Al2S3 reacted = 0.0204 mol.
So, moles of Al2S3 remaining = (0.6406 mol - 0.0204 mol) = 0.6202 mol.

To convert moles to grams, you can use the molar mass of Al2S3, which is 150.16 g/mol.

Mass of Al2S3 remaining = (0.6202 mol) * (150.16 g/mol) = 93.152 g.

Therefore, approximately 93.152 grams of solid Al2S3 remain after the reaction.