Respuesta :
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 %
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
