Ask Question
30 January, 17:52

How many grams of Aldol product can be produced from the complete reaction of 0.2 grams of vanillin with an excess of acetone in the presence of aqueous base? Enter only the number with two significant figures.

+3
Answers (1)
  1. 30 January, 21:29
    0
    There is 0.25 grams of C11H12O3 produced

    Explanation:

    Step1: Data given

    vanillin = C8H8O3

    Mass of vanillin = 0.2 grams

    Molar mass of vanillin = 152.15 g/mol

    Acetone = 58.08 g/mol

    Step 2: The balanced equation

    C8H8O3 + C3H6O → H2O + C11H12O3

    Step 3: Calculate moles of C8H8O3

    Moles C8H8O3 = mass / molar mass

    Moles C8H8O3 = 0.2 grams / 152.15 g/mol

    Moles C8H8O3 = 0.0013 moles

    Step 4: Calculate moles of C11H12O3

    For 1 mol vanillin we need 1 mol acetone to produce 1 mol C11H12O3

    Step 5: Calculate mass of C11H12O3

    Mass C11H12O3 = moles * molar mass

    Mass C11H12O3 = 0.0013 moles * 192.21 g/mol

    Mass C11H12O3 = 0.25 grams

    There is 0.25 grams of C11H12O3 produced
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “How many grams of Aldol product can be produced from the complete reaction of 0.2 grams of vanillin with an excess of acetone in the ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers