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

Question

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

Answer 1

Al₂S₃ + 6H₂O ==> 2Al(OH)₃ + 3H₂S .. balanced equation

molar mass Al₂S₃ = 150.16 g

molar mass H₂O = 18.015 g

One easy way to find the limiting reactant is to divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant. In the current problem we have the following:

Al₂S₃: 15.00 g x 1 mol / 150.16 g = 0.09989 mols (÷1->0.0999)

H₂O: 10.00 g x 1 mol / 18.015 g = 0.5551 mols (÷6->0.0925)

Since 0.0925 is less than 0.0999, H₂O is the limiting reactant. This makes Al₂S₃ the excess reactant.

To find out how much Al₂S₃ is left over, we will use the stoichiometry of the balanced equation to first find out how much Al₂S₃ was used up, and then subtract that from the original amount present.

Al₂S₃ used up: 0.5551 mols H₂O x 1 mol Al₂S₃ / 6 mol H₂O = 0.09252 mols Al₂S₃ used

Al₂S₃ left over: 0.09989 mols - 0.09252 mols = 0.007370 Al₂S₃ mols left over

0.007370 mols Al₂S₃ x 150.16 g / mol = 1.107 g Al₂S₃ left over