15.00 g aluminum sulfide and 10.00 g water react until the limiting reagent is used up. How much of the excess reactant is left over? Al2S3 + 6 H2O → 2Al(OH)3 + 3 H2S (Show your work )

Answer 1

Al₂S₃ + 6H₂O ==> 2Al(OH)₃ + 3H₂S ... BALANCED EQUATION

An easy way to identify the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation and whichever value is less indicates the limiting reactant.

Al₂S₃: 15.00 g Al₂S₃ x 1 mol Al₂S₃ / 150.2 g = 0.09987 mols Al₂S₃ (÷1->0.0999)

H₂O: 10.00 g H2O x 1 mol H₂O / 18.02 g = 0.5549 mols H₂O (÷6->0.0925)

Since 0.0925 is less than 0.0999, H₂O is the limiting reactant.

To find how much excess reactant (Al₂S₃) is left over, we use the stoichiometry of the balanced equation to first find out how much has reacted. Then subtract that from the amount originally present to find the amount left over,

Al₂S₃ used up: 0.5549 mols H₂O x 1 mol Al₂S₃ / 6 mol H₂O = 0.0925 mols Al₂S₃ used

Al₂S₃ remaining: 0.09987 mols - 0.0925 mols = 0.00737 Al₂S₃ mols left over

If you want this in grams: 0.00737 mols x 150.2 g/mol = 1.107 g Al₂S₃ left over