Ask Question
9 September, 03:27

A.) How many moles of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate? N=?

b.) How many grams of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate with H2? N=?

c.) How many grams of ethyl alcohol are produced by reaction of 16.0 g of ethyl acetate with H2? N=?

D.) How many grams of H2 are needed to react with 16.0 g of ethyl acetate? N=?

+4
Answers (1)
  1. 9 September, 04:47
    0
    a) 8.4 moles ethyl alcohol

    b) 387 grams ethyl alcohol

    c) 16.77 grams ethyl alcohol

    d) 0.735 grams H2

    Explanation:

    Step 1: Data given

    Molar mass of ethyl acetate = 88.11 g/mol (C4H8O2)

    Molar mass of ethyl alcohol = 46.07 g/mol (C2H6O)

    Step 2: The balanced equation

    C4H8O2 + 2H2 → 2C2H6O

    Step 3: How many moles of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate?

    For 1 mol mol ethyl acetate we need 2 moles H2 to produce 2 moles ethyl alcohol.

    For 4.2 mol of ethyl acetate we'll have 2*4.2 = 8.4 moles ethyl alcohol

    Step 4: How many grams of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate with H2?

    Mass of ethyl alcohol = moles * molar mass

    Mass ethyl alcohol = 8.4 moles * 46.07 g/mol

    Mass ethyl alcohol = 387 grams

    Step 5: How many grams of ethyl alcohol are produced by reaction of 16.0 g of ethyl acetate with H2?

    Moles ethyl acetate = 16.0 grams / 88.11 g/mol = 0.182 moles

    Moles ethyl alcohol = 2*0.182 = 0.364 moles

    Mass ethyl alcohol = 16.77 grams

    Step 6: How many grams of H2 are needed to react with 16.0 g of ethyl acetate?

    For 0.182 moles ethyl acetate we need 2*0.182 = 0.364 moles H2

    0.364 moles H2 = 0.364 * 2/02 g/mol = 0.735 grams H2
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A.) How many moles of ethyl alcohol are produced by reaction of 4.2 mol of ethyl acetate? N=? b.) How many grams of ethyl alcohol are ...” 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