Answer:
There is plenty of oxygen to react with all of the octane. Excess oxgen will be 14.8 grams.
Step-by-step explanation:
The first thing we need is to balance the chemical equation. We'll start with just the chemical formulas and then adjust their coefficients until the equation is balanced.
The combustion of octane:
C8H18 + O2 ⇒ CO2 + H2O
Each octane molecule provides 8 carbons and 18 hydrogens. It is the most comples molecule, so start with it by assigning a coefficient of 1 and find "homes" for all 8 carbons.
1C8H18 + O2 ⇒ CO2 + H2O
The only place carbons appear in the products is in the CO2, so lets give it a coefficient of 8:
1C8H18 + O2 ⇒ 8CO2 + H2O
Now lets find homes for the 18 hydrogen atoms. The only place hydrogen shows up in a product is the H2O, so assign it a coefficient of 9:
1C8H18 + O2 ⇒ 8CO2 + 9H2O
We've balanced the atoms from the octane, but it is clear we'll need a lot more oxygen atoms. There are a total of (16+9 = 25) 25 oxygen atoms in the 8CO2 + 9H2O product, The oinly place we can get oxygen is from the O2. We need 12.5 O2 molecules for 25 O atoms.
1C8H18 + 12.5O2 ⇒ 8CO2 + 9H2O
The equation is balanced, but there is no such thing as a fraction of an atom. So lets multiply everything by 2 to obtain whole coefficients:
2C8H18 + 25O2 ⇒ 16CO2 + 18H2O
The equation is now balanced. Despite the awkward numbers, it is the one we muct use to determine how much octane is left over.
Calculate the number of moles contained in the given masses of the reactants:
The molar mass of each molecule is needed:
Octane: 114 g/mole
O2 = 32 grams/mole
Calculate the number of moles of each:
Octane: 8.00g/(114 g/mole) = 0.07 moles octane
O2: 43.0 g/(32 g/mole) = 1.34 moles O2
The balanced equation tells us that we need 12.5 moles O2 for every 1 mole of octane. That is a molar ratio of (12.5 mole O2)/(1 mole octane).
Since we started with 0.07 moles octane, we'll only consume
(0.07 moles octane)*(12.5 mole O2)/(1 mole octane) = 0.88 moles O2
We start with 1.34 moles of O2, so there is plenty for the reaction.
Although the problem infers that it is the octane that should be left over, this tells us that there is plaenty of oxygen, and that it is the oxygen that is in excess.
The excess oxygen is (1.34 moles O2) - (0.88 moles O2 consumed) = 0.47 moles O2 that is unreacted.
At 32 g/moles, this means we'll have (0.47 moles)*(32g/mole) = 14.8 grams of O2 remains unreacted.