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
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Answers ( 2 )
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:
Stoichiometrically from the equation,
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,
Hence, the 112.64 gram is hypothetically yield, and actual yield prodced is 28.16 gram. Now, calculate the percentage yield, we get:
Therefore, 25% of octane will be yielded.
To know more about combustion reaction and percentage yield, refer to the following link:
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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 %