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26 October, 19:40

During the following chemical reaction, 46.3 grams of C3H6O react with 73.2 grams of O2

(Balanced Equation) C3H6O + 4 O2 = 3 CO2 + 3 H2O

(Questions)

a. What is the limiting reactant?

b. How many grams of CO2 are produced?

c. How much of the excess reactant remains?

d. If the percent yield of CO2 is 45.3%, what is the actual yield?

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Answers (1)
  1. 26 October, 22:38
    0
    a) O2 is the limiting reactant

    b) 75.70 grams CO2 (theoretical yield)

    c) There remains 12.81 grams of C3H6O

    d) The actual yield CO2 is 34.29 grams

    Explanation:

    Step 1: Data given

    Mass of C3H6O = 46.3 grams

    Mass of O2 = 73.2 grams

    Molar mass of C3H6O = 58.08 g/mol

    Molar mass of O2 = 32 g/mol

    Step 2: The balanced equation

    C3H6O + 4O2 → 3 CO2 + 3H2O

    Step 3: Calculate moles C3H6O

    Moles C3H6O = mass C3H6O / molar mass C3H6O

    Moles C3H6O = 46.3 grams / 58.08 g/mol

    Moles C3H6O = 0.793 moles

    Step 4: Calculate moles O2

    Moles O2 = 73.2 grams / 32 g/mol

    Moles O2 = 2.29 moles

    Step 5: Calculate limiting reactant

    For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

    O2 is the limiting reactant. It will completely be consumed. (2.29 moles).

    C3H6O is in excess. There will react 2.29/4 = 0.5725 moles C3H6O

    There will remain 0.793 - 0.5725 = 0.2205 moles C3H6O

    This is 0.2205 moles * 58.08 g/mol = 12.81 grams

    Step 6: Calculate moles of CO2

    For 1 mol C3H6O we need 4 moles of O2 to produce 3 moles CO2 and 3 moles H2O

    For 2.29 moles O2 we need 3/4 * 2.29 = 1.72 moles CO2

    This is 1,72 moles * 44.01 g/mol = 75.70 grams CO2

    Step 7: Calculate actual yield

    % yield = 45.3 % = 0.453 = (actual yield / theoretical yield)

    actual yield = 0.453 * 75.70 = 34.29 grams
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