The given question is incomplete. the complete question is:
The world burns the fossil fuel equivalent of approximately 
kg of petroleum per year. Assume that all of this petroleum is in the form of octane. Calculate how much CO2 in kilograms is produced by world fossil fuel combustion per year.( Hint: Begin by writing a balanced equation for the combustion of octane.)
Answer: 
Explanation:
Combustion is a chemical reaction in which hydrocarbons are burnt in the presence of oxygen to give carbon dioxide and water.
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.
To calculate the moles :
According to stoichiometry :
As 2 moles of octane give = 16 moles of 
Thus 
of octane give =
of
Mass of 
Thus of is produced by world fossil fuel combustion per year.
Here, we use the mole as we would use any other collective number: a dozen eggs; a Bakers' dozen; a Botany Bay dozen.
Of course, the mole specifies a much larger quantity, and if I have a mole of stuff then I have
6.022
×
10
23
individual items of that stuff. We can also specify an equivalent mass, because we also know the mass of a mole of iron, and a mole of oxygen etc........The mole is thus the link between the macro world of grams and kilograms and litres, that which we can measure out in the lab, to the micro world of atoms, and molecules, that which we can perceive only indirectly.
Here we have the formula unit
F
e
2
(
S
O
4
)
3
. If there is a mole of formula units, there are necessarily 2 moles of iron atoms, 3 sulfate ions,.......etc.
The balanced chemical reaction would be as follows:
<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10
We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:
8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2
Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:
0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
Answer: <u><em>True</em></u>
Explanation:
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