combustion reaction

Question

When 11.8 g of C₈H₁₈ underwent a combustion reaction with 8.84 g of O₂, it produced 1.15 g of H₂O. 

Balance the reaction and answer the following questions. 

What is the mass (in g) of remaining (unreacted) octane (C8H18)? 

Answer 1

Because it is 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.

1st we must balance the equation:

Balance

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O (not taking into account the combustion of air)

We are told 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)