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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

Respuesta :

transitionstate
The combustion reaction of octane is as follow,
                           2 C₈H₁₈  +  25 O₂     →     16 CO₂  +  18 H₂O

First find out the limiting reagent,
As,

2 moles Octane required  =  25 moles of O₂
So,

4 moles will require  =  X moles of O₂

Solving for X,
                        X  =  (25 mol × 4 mol) ÷ 2 mol

                        X  =  50 mol of O₂

So, 4 moles of Octane will require 50 moles of Oxygen. But we are provided with only 4 moles of O₂. Hence, O₂ is the limiting reagent and will control the yield of CO₂. So,

25 mol O₂ produced  =  704 g (16 mol) of CO₂

Then,

4 mol of O₂ will produce  =  X g of CO₂

Solving for X,
                        X  =  (4 mol × 704 g) ÷ 25 mol

                        X  =  112.64 g of CO₂

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 / 112.64 × 100

                    %age Yield  =  25 %

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 C₈H₁₈ + 25 O₂   →   16 CO₂ + 18 H₂O

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 O₂.

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 O₂ produced  =  704 g (16 mol) of CO₂
  • 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

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