Ask Question
5 November, 08:11

In the organic combustion reaction of 41.9 g of octane (C8H18) with excess oxygen, what volume (in L) of carbon dioxide is produced if the reaction is performed at STP?

+2
Answers (1)
  1. 5 November, 08:36
    0
    The volume CO2 produced is 65.8 L

    Explanation:

    Step 1: Data given

    Mass of octane = 41.9 grams

    Molar mass octane = 114.23 g/mol

    Step 2: The balanced equation

    2C8H18 + 25O2 → 16CO2 + 18H2O

    Step 3: Calculate moles octane

    Moles octane = mass octane / molar mass octane

    Moles octane = 41.9 grams / 114.23 g/mol

    Moles octane = 0.367 moles

    Step 4: Calculate moles CO2

    For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

    For 0.367 moles octane we need 8*0.367 = 2.936 moles

    Step 5: Calculate volume of CO2

    1 mol = 22.4 L

    2.936 moles = 22.4 * 2.936 = 65.8 L

    The volume CO2 produced is 65.8 L
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “In the organic combustion reaction of 41.9 g of octane (C8H18) with excess oxygen, what volume (in L) of carbon dioxide is produced if the ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers