## 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

in progress 0

1. 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. 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 %