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

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To determine the mass of CO2 produced from the combustion of 15g of C8H18 with 15g of oxygen at a 93% yield, first calculate the moles of octane in 15g. The balanced reaction shows that 1 mole of C8H18 produces 8 moles of CO2. If the reaction had a 100% yield, calculate the moles of CO2 produced and then apply the 93% yield to find the actual amount. Finally, convert the moles of CO2 to mass using its molecular weight. This approach will yield the mass of CO2 produced in the reaction.
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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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