Ask Question
31 May, 08:59

A reaction was conducted between barium nitrate and sodium phosphate. 3 Ba (NO3) 2 + 2 Na3PO4 → 6 NaNO3 + Ba3 (PO4) 2 What is the percent yield if 0.3 mol Ba (NO3) 2 and 0.25 mol Na3PO4 react to produce 0.095 mol Ba3 (PO4) 2?

+2
Answers (1)
  1. 31 May, 10:57
    0
    The % yield is 95.0 %

    Explanation:

    Step 1: Data given

    Number of moles Ba (NO3) 2 = 0.3 moles

    Number of moles Na3PO4 = 0.25 moles

    Number of Ba3 (PO4) 2 = 0.095 moles

    Step 2: The balanced equation

    3 Ba (NO3) 2 + 2 Na3PO4 → 6 NaNO3 + Ba3 (PO4) 2

    Step 3: Calculate the limiting reactant

    For 3 moles Ba (NO3) 2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3 (PO4) 2

    Ba (NO3) 2 is the limiting reactant. It will compeltely be consumed completely (0.3 moles). Na3PO4 is in excess. There will react 0.20 moles.

    There will remain 0.25-0.20 = 0.05 moles

    Step 4: Calculate moles Ba3 (PO4) 2

    For 3 moles Ba (NO3) 2 we need 2 moles Na3PO4 to produce 6 moles NaNO3 and 1 mol Ba3 (PO4) 2

    For 0.3 moles Ba (NO3) 2 we'll have 0.3/3 = 0.1 mol Ba3 (PO4) 2

    Step 5: Calculate percent yield

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

    % yield = (0.095 / 0.1) * 100 %

    % yield = 95.0 %

    The % yield is 95.0 %
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A reaction was conducted between barium nitrate and sodium phosphate. 3 Ba (NO3) 2 + 2 Na3PO4 → 6 NaNO3 + Ba3 (PO4) 2 What is the percent ...” 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