Give the percent yield when 28.16 g of co2 are formed from the reaction of 4.000 moles of c8h18 with 4.000 moles of o2. 2 c8h18(l) + 25 o2(g

Question

Give the percent yield when 28.16 g of co2 are formed from the reaction of 4.000 moles of c8h18 with 4.000 moles of o2. 2 c8h18(l) + 25 o2(g) → 16 co2(g) + 18 h2o(g) molar mass co2 = 44.01 g/mol

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2022-04-10T18:43:26+00:00 2 Answers 0

Answers ( 2 )

  1. Eva
    0
    2022-04-10T18:44:56+00:00
    Reply

    A combustion reaction is known as the reaction in which a fuel or substance is heated in the presence of oxygen or oxide. The percentage yield of octane in the combustion reaction is 25 %.

    The chemical reaction between octane and oxygen is as follows:

    • 2 Cu2088Hu2081u2088 + 25 Ou2082 u2192 16 COu2082 + 18 Hu2082O

    Stoichiometrically from the equation,

    • 2 moles of octane = 25 moles of oxygen
    • 4 moles of octane = x moles of oxygen
    • [tex]text x&= dfrac{(25 text{mol} times 4 text{mol})} {2}[/tex]
    • x = 50 moles of oxygen or Ou2082.

    For every 4 moles of octane, 50 moles of oxygen will be produced. Since, oxygen is the limiting reagent, it will control the yield of carbon dioxide.

    Now,

    • 25 mol Ou2082 produced = 704 g (16 mol) of COu2082n
    • 4 moles of oxygen will produce = x g of carbon dioxide
    • Solving for x:
    • [tex]text x&= dfrac{(4 text{mol} times 704 text{mol})} {25}[/tex]
    • x = 112.64 gram of carbon dioxide

    Hence, the 112.64 gram is hypothetically yield, and actual yield prodced is 28.16 gram. Now, calculate the percentage yield, we get:

    • Percentage yield = [tex]dfrac {28.16}{112.64}times 100[/tex]
    • Percentage yield = 25%

    Therefore, 25% of octane will be yielded.

    To know more about combustion reaction and percentage yield, refer to the following link:

    https://brainly.com/question/13873164?referrer=searchResults

  2. Morgan
    0
    2022-04-10T18:45:25+00:00
    Reply

    The combustion reaction of octane is as follow,
    2 Cu2088Hu2081u2088 + 25 Ou2082 u2192 16 COu2082 + 18 Hu2082O

    First find out the limiting reagent,
    As,

    2 moles Octane required = 25 moles of Ou2082
    So,

    4 moles will require = X moles of Ou2082

    Solving for X,
    X = (25 molx 4 mol)u00f7 2 mol

    X = 50 mol of Ou2082

    So, 4 moles of Octane will require 50 moles of Oxygen. But we are provided with only 4 moles of Ou2082. Hence, Ou2082 is the limiting reagent and will control the yield of COu2082. So,

    25 mol Ou2082 produced = 704 g (16 mol) of COu2082

    Then,

    4 mol of Ou2082 will produce = X g of COu2082

    Solving for X,
    X = (4 molx 704 g)u00f7 25 mol

    X = 112.64 g of COu2082

    Hence,112.64 g is the theoretical yield. And Actual yield given is 28.16 g. So, %age yield is calculated as,

    %age Yield = 28.16 / 112.64x 100

    %age Yield = 25 %

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