15.00 g aluminum sulfide and 10.00 g of water react until the limiting reactant is used up the products for the reaction are aluminum hydroxide and hydrogen sulfide gas

Question

A. Calculate the theoretical yield of H2S

B. A student performs the reaction. The student's empty flask has a mass of 28.55 grams. The flask and H2S have a combined mass of 36.77 grams. Determine the % yield of the reaction.

C. Determine the mass of excess reactant remaining at the end of the reaction

Answer 1

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

moles Al₂S₃ present = 15.00 g x 1 mole/150.16 g = 0.09989 moles

moles H₂O present = 10.00 g x 1 mole/18.02 g = 0.5549 moles

Limiting reactant = H₂O since it requires 6 moles H₂O per 1 mole Al₂S₃ and there isn't enough.

A. Theoretical yield of H₂S: 0.5549 mol H₂O x 3 mol H₂S/6 mol H₂O x 34.1 g/mol = 9.461 g H₂S

B. % yield: mass H₂S collected = 36.77 g - 28.55 g = 8.220 g.

% yield = actual yield/theoretical yield (x100%) = 8.220 g/9.461 g (x100%) = 86.88% yield

C. Mass of excess reactant: 0.5549 mol H₂O x 1 mol Al₂S₃/6 mol H₂O = 0.09248 moles Al₂S₃ used up

0.09989 mol - 0.09248 mol = 0.007410 moles Al₂S₃ remaining

0.007410 moles x 150.16 g/mole = 1.113 g Al₂S₃ remaining