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1 July, 17:24

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

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  1. 1 July, 17:46
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    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%
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