It is given that its a combustion reaction, we know that the ending compounds are CO2 and H2O alone. To get this, we add O2 in a combustion reaction to our initial compound.
C8H18 + O2 ---> CO2 + H2O ( + trace amounts of C8H18 left) We are asked about the C8H18 left meaning we have a limiting reactant and it is not completely water and CO2 left due to the limiting reagent. So we are trying to find the excess reactant.
To balance the equation we must balance all molecules ;
2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O (not taking into account the combustion of air)
It is given that it is incomplete combustion so we must use the grams of product to calculate the amount of reactant used due to the limiting reagent
1.15 g H2O * ( 1 mol H2O / 18.01532 g H2O) *( 2 mol C8H18 / 18 mol H2O) * ( 1 mol C8H18 / 114.24 g C8H18) = 0.00111755 = 0.00118 g C8H18 reacted
11.8 g C8H18 in the initial reaction, 0.0118 g reacted to get 1.15 g H2O
11.8-0.00118 = 11.79882 = 11.8 g C8H18 left in the balanced equation (meaning almost none of it was reacted)
Learn more about balance equation reaction ; https://brainly.com/question/11322377?referrer=searchResults
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