How Much CO2 is Produced in a Combustion Reaction with 93% Yield?

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SUMMARY

The combustion reaction of octane (C8H18) produces water (H2O) and carbon dioxide (CO2) in a stoichiometric ratio. Given 15g of C8H18 and 15g of oxygen, the theoretical yield of CO2 can be calculated based on the molar mass of octane and the reaction's stoichiometry. With a 93% yield, the actual mass of CO2 produced can be determined by first calculating the moles of octane and then applying the yield percentage to the theoretical mass of CO2 generated from complete combustion.

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Hi, I'm stuck with this problem.

C8H18 ------> H20 + 8CO2

There is 15g of C8H18 and 15g of Oxygen to make combustion reaction.

If the actual yield percent of CO2 is 93%, what mass of CO2 will be produced?

I approched with Morecular weight to get mol. and...I totally get lost.
 
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I would usually start here by asking you to balance your equation but you don't need to in this case. You notice that there is a particular ratio of CO2 to octane in this example? Find the number of moles of octane you have in 15 g. How many moles of CO2 will be produced if the reaction has a 100% yield? A 93% yield? From moles of CO2 you should be able to determine mass...
 

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