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

To find the limiting reactant, first find moles of each reactant present:

moles Al₂S₃ = 15.00 g x 1 mol/150.2 g = 0.09987 moles Al₂S₃

moles H₂O = 10.00 g x 1 mol/18.02 g = 0.5549 moles H₂O

Dividing each by the respective coefficient we have...

-Al₂S₃ = 0.09987/1 = 0.09987

-H₂O = 0.5549/6 = 0.0925 <= H₂O will be LIMITING

A. Theoretical yield of H₂S:

0.5549 moles H₂O x 3 moles H₂S/6 moles H₂O x 34.1 g/mole = 9.461 g H₂S

B. % yield of H₂S:

36.77 g - 28.55 g = 8.22 g H₂S recovered

% yield = actual yield/theoretical yield (x100) = 8.22g/9.461 g (x100) = 86.9% yield

C. Excess reactant is Al₂S₃.

0.5549 moles H₂O x 1 mol Al₂S₃/6 moles H₂O = 0.09248 moles Al₂S₃ used up

moles Al₂S₃ remaining = 0.09987 moles - 0.09248 moles = 0.00739 moles remaining

mass of Al₂S₃ remaining = 0.00739 moles x 150.2 g/mole = 1.110 g