The octane in gasoline burns according to the following equation. $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}+25 \mathrm{O}_{2} \longrightarrow 16 \mathrm{CO}_{2}+18 \mathrm{H}_{2} \mathrm{O} $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to react fully with \(6.84 \mathrm{~mol}\) of octane? (b) How many moles of \(\mathrm{CO}_{2}\) can be formed from 0.511 mol of octane? (c) How many moles of water are produced by the combustion of \(8.20 \mathrm{~mol}\) of octane? (d) If this reaction is used to synthesize \(6.00 \mathrm{~mol}\) of \(\mathrm{CO}_{2}\), how many moles of oxygen are needed? How many moles of octane?

Understanding the mole-to-mole ratio is crucial when working with chemical reactions. In stoichiometry, the mole-to-mole ratio is derived from the coefficients in a balanced chemical equation. It indicates the relative amounts of reactants and products involved in the reaction.

Let's use the combustion of octane as an example, where we have a balanced equation:
\[2 \mathrm{C}_{8} \mathrm{H}_{18} + 25 \mathrm{O}_{2} \longrightarrow 16 \mathrm{CO}_{2} + 18 \mathrm{H}_{2} \mathrm{O}\].
In this equation, the coefficients suggest that 2 moles of octane (\(\mathrm{C}_{8}\mathrm{H}_{18}\)) react with 25 moles of oxygen (\(\mathrm{O}_{2}\)) to produce 16 moles of carbon dioxide (\(\mathrm{CO}_{2}\)) and 18 moles of water (\(\mathrm{H}_{2}\mathrm{O}\)). Therefore, the mole-to-mole ratio between octane and oxygen is 2:25, octane to carbon dioxide is 2:16, and octane to water is 2:18.

When solving stoichiometric problems, these ratios enable us to convert between amounts of different substances involved. For instance, if we begin with 6.84 moles of octane, we could use the mole-to-mole ratio to find out how much oxygen is required or how much water and carbon dioxide would be produced.

The proper balancing of a chemical equation is the foundation of stoichiometry. A chemical equation must be balanced to obey the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

Every element must have the same number of atoms on both the reactant and product sides of the equation. For the combustion of octane, the balanced equation is: \[2 \mathrm{C}_{8} \mathrm{H}_{18} + 25 \mathrm{O}_{2} \longrightarrow 16 \mathrm{CO}_{2} + 18 \mathrm{H}_{2} \mathrm{O}\].
To balance a chemical equation, one must adjust the coefficients—the numbers placed before the chemical formulas—without altering the formulas themselves.

For educators, reinforcing the importance of balancing equations cannot be overstated. It enables students to predict the outcomes of reactions accurately. Balancing is not just a mathematical exercise; it represents the actual stoichiometry of the reaction and is critical to quantitative analysis in chemistry.

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of light or heat. These reactions are exothermic, meaning they produce energy. Combustion of hydrocarbons, like octane in gasoline, is a common example and can be represented by a general equation:

\[\mathrm{Hydrocarbon} + \mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\].
The products of complete combustion are typically carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)), assuming a plentiful supply of oxygen. In the case of octane, the reaction is particularly relevant to the automotive industry as it underpins the operation of internal combustion engines.

Students must learn to identify and solve for the various components of a combustion reaction, which includes being able to calculate the amounts of reactants needed and the amounts of products formed using mole-to-mole ratios—all of which are garnered from a balanced chemical equation.