Answer:
Limiting reagent: barium nitrate
Theoretical yield: 2.29 g BaS
Percent yield: 87%
Explanation:
The corrected balanced reaction equation is:
K₂S + Ba(NO₃)₂ ⇒ 2KNO₃ + BaS
The amount of potassium sulfide added is:
(25.0 mL)(1.20mol/L) = 30 mmol
The amount of barium nitrate added is:
(15.0mL)(0.900mol/L = 13.5 mmol
Since the reactant react in a 1:1 molar ratio, barium nitrate is the limiting reagent. If all of the limiting reagent reacts, the amount of barium sulfide produced is:
(13.5 mmol Ba(NO₃)₂)(BaS/Ba(NO₃)₂ ) = 13.5 mmol BaS
Converting this amount to grams gives the theoretical yield of BaS (molar mass 169.39 g/mol).
(13.5 mmol)(169.39 g/mol)(1g/1000mg) = 2.29 g BaS
The percent yield is calculated as follows:
(actual yield) / (theoretical yield) x 100%
(2.0 g) / (2.29 g) x 100% = 87%
