Chemistry, 21.07.2019 15:00 Dmoney7784

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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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 %

Answer from: sk9600930

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 $\text x&= \dfrac{(25 \text{mol} \times 4 \text{mol})} {2}$ 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 dioxideSolving for x: $\text x&= \dfrac{(4 \text{mol} \times 704 \text{mol})} {25}$ 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 = $\dfrac {28.16}{112.64}\times 100$ Percentage yield = 25%

Therefore, 25% of octane will be yielded.

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

link

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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...

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