Ask Question
5 May, 22:20

A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and this precipitation reaction occurs: K2SO4 (aq) + Ba (NO3) 2 (aq) →BaSO4 (s) + 2KNO3 (aq) K2SO4 (aq) + Ba (NO3) 2 (aq) →BaSO4 (s) + 2KNO3 (aq) The solid BaSO4BaSO4 is collected, dried, and found to have a mass of 2.57 gg. Determine the limiting reactant, the theoretical yield, and the percent yield.

+3
Answers (1)
  1. 5 May, 22:31
    0
    The limiting reactant is Ba (NO3) 2

    The theoretical yield BaSO4 is 2.97 grams

    The percent yield of the reaction is 86.5 %

    Explanation:

    Step 1: Data given

    Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

    Volume of a 0.860 M barium nitrate solution (Ba (NO3) 2 = 14.8 mL = 0.0148 L

    The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

    Step 2: The balanced equation

    K2SO4 (aq) + Ba (NO3) 2 (aq) → BaSO4 (s) + 2KNO3 (aq)

    Step 3: Calculate moles

    Moles = volume * molarity

    Moles K2SO4 = 0.022 L * 1.16 M

    Moles K2SO4 = 0.02552 moles

    Moles Ba (NO3) 2 = 0.0148 L * 0.860 M

    Moles Ba (NO3) 2 = 0.012728 moles

    Step 4: Calculate the limiting reactant

    For 1 mol K2SO4 we need 1 mol Ba (NO3) 2 to produce 1 mol BaSO4 and 2 moles KNO3

    Ba (NO3) 2 is the limiting reactant. It will completely be consumed. (0.012728 moles). K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

    Step 5: Calculate moles BaSO4

    ‬For 1 mol K2SO4 we need 1 mol Ba (NO3) 2 to produce 1 mol BaSO4 and 2 moles KNO3

    For 0.012728 moles Ba (NO3) 2 we'll have 0.012728 moles BaSO4

    Step 6: Calculate mass BaSO4

    Mass BasO4 = moles BaSO4 * molar mass BaSO4

    Mass BaSO4 = 0.012728 moles * 233.38 g/mol

    Mass BaSO4 = 2.97 grams

    Step 7: Calculate the percent yield

    % yield = (actual yield / theoretical yield) * 100 %

    % yield = (2.57 grams / 2.97 grams) * 100 %

    % yield = 86.5 %

    The limiting reactant is Ba (NO3) 2

    The theoretical yield BaSO4 is 2.97 grams

    The percent yield of the reaction is 86.5 %
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and this ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers