If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reaction occurs. K25 (aq) + Ba (NO₃), (ag) 2 KNO₂ (aq) + Bas (s). If the above reaction produced 2. og Bas determine the limiting reactant, theoretical field and percent yield, Aunting reactant - theoetical yeeld; % yield - Show All Work on seperate Sheets

Question

Charlize

Chemistry

31 December, 02:42

If you mix a 25.0mL sample of a 1.20m potassium sulfide solution with 15.0 mL of a 0.900m basium nitrate solution and the reaction occurs. K25 (aq) + Ba (NO₃), (ag) 2 KNO₂ (aq) + Bas (s). If the above reaction produced 2. og Bas determine the limiting reactant, theoretical field and percent yield, Aunting reactant - theoetical yeeld; % yield - Show All Work on seperate Sheets

+1

Answers (1)

Know the Answer?

Jaelynn Dunlap · Answer 1

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%