Ask Question
5 August, 18:50

If 2.00 grams of CO2 were produced when 4.00 grams of C5H12 were combusted in excessoxygen, what is the percent yield of the combustion reaction?

+5
Answers (1)
  1. 5 August, 19:19
    0
    The % yield of the reaction is 16.4 %

    Explanation:

    Step 1: Data given

    Mass of CO2 = 2.00 grams

    Mass of C5H12 = 4.00 grams

    O2 is in excess

    Molar mass of CO2 = 44.01 g/mol

    Molar mass of C5H12 = 72.15 g/mol

    Molar mass of O2 = 32 g/mol

    Step 2: The balanced equation

    C5H12 + 8O2 → 5CO2 + 6H2O

    Step 3: Calculate moles C5H12

    Moles C5H12 = mass C5H12 / molar mass C5H12

    Moles C5H12 = 4.00 grams / 72.15 g/mol

    Moles C5H12 = 0.0554 moles

    Step 4: Calculate moles of CO2

    For 1 mol pentane, we need 8 moles of O2 to produce 5 moles CO2 and 6 moles H2O

    For 0.0554 moles C5H12 we'll have 5*0.0554 = 0.277 moles CO2

    Step 5: Calculate mass CO2

    Mass CO2 = moles CO2 * molar mass CO2

    Mass CO2 = 0.277 moles ¨44.01 g/mol

    Mass CO2 = 12.2 grams CO2

    Step 6: Calculate % yield

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

    % yield = (2.00 grams / 12.2 grams) * 100%

    % yield = 16.4 %

    The % yield of the reaction is 16.4 %
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “If 2.00 grams of CO2 were produced when 4.00 grams of C5H12 were combusted in excessoxygen, what is the percent yield of the combustion ...” 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