Question

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the main component of gasoline, proceeds as follows: \(2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g)\) (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.50 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn \(15.0 \mathrm{gal}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) (the capacity of an average fuel tank)? (d) How many grams of \(\mathrm{CO}_{2}\) are produced when 15.0 gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\) are combusted?

Short answer

(a) To burn 1.50 moles of C₈H₁₈, we need \(18.75\ \text{moles of O₂}\). (b) To burn 10.0 g of C₈H₁₈, we need \(87.7\ \text{grams of O₂}\). (c) To burn 15.0 gal of C₈H₁₈, we need \(3,876,000\ \text{grams of O₂}\). (d) When burning 15.0 gal of C₈H₁₈, we produce \(6,213,000\ \text{grams of CO₂}\).

Step-by-step solution

(a) Moles of O₂ needed for 1.50 mol of C₈H₁₈)

For this part, we will use stoichiometry to relate moles of C₈H₁₈ and O₂ from the balanced combustion equation. From the balanced equation, $$2\ \mathrm{C}_{8}\mathrm{H}_{18}(l)+25\ \mathrm{O}_{2}(g)\longrightarrow16\ \mathrm{CO}_{2}(g)+18\ \mathrm{H}_{2}\mathrm{O}(g)$$ To find the moles of O₂ needed to burn 1.50 moles of C₈H₁₈, set up a stoichiometric ratio between moles of O₂ and moles of C₈H₁₈: $$\frac{25\ \text{moles of O₂}}{2\ \text{moles of C₈H₁₈}}=\frac{x\ \text{moles of O₂}}{1.50\ \text{moles of C₈H₁₈}}$$ Next, solve for x. (b) Grams of O₂ needed to burn 10.0 g of C₈H₁₈

Find moles of C₈H₁₈ from given mass

We can use the molar mass of C₈H₁₈, which is \(8(12.01 \ \mathrm{g/mol}) + 18(1.008 \ \mathrm{g/mol}) = 114.23 \ \mathrm{g/mol}\). Now convert 10.0 g of C₈H₁₈ to moles: $$\text{moles of C₈H₁₈}=\frac{10.0\ \mathrm{g}}{114.23\ \mathrm{g/mol}}$$

Find moles of O₂ needed using stoichiometry

Use the balanced equation and the ratio from part (a) to find the moles of O₂ needed for the moles of C₈H₁₈ found in step 1.

Convert moles of O₂ to grams

We can use the molar mass of O₂, which is \(2(16.00 \ \mathrm{g/mol}) = 32.00 \ \mathrm{g/mol}\). Now convert the moles of O₂ from step 2 to grams of O₂: $$\text{grams of O₂}=\text{moles of O₂}\times32.00\ \mathrm{g/mol}$$ (c) Grams of O₂ required to burn 15.0 gal of C₈H₁₈

Convert gallons to grams of C₈H₁₈

We can use the density of octane and convert 15.0 gal to grams: $$15.0\ \mathrm{gal} \times \frac{3.785 \ \mathrm{L}}{1 \ \mathrm{gal}}\times \frac{1000 \ \mathrm{mL}}{1 \ \mathrm{L}} \times \frac{0.692 \ \mathrm{g}}{1 \ \mathrm{mL}}$$

Find moles of C₈H₁₈ from grams

Use the molar mass of C₈H₁₈ found in part (b) to convert the grams of C₈H₁₈ to moles.

Find moles of O₂ needed using stoichiometry

Use the balanced equation and the ratio from part (a) to find the moles of O₂ needed for the moles of C₈H₁₈ found in step 2.

Convert moles of O₂ to grams

Use the molar mass of O₂ found in part (b) to convert the moles of O₂ from step 3 to grams of O₂. (d) Grams of CO₂ produced when 15.0 gal of C₈H₁₈ are combusted

Use grams or moles of C₈H₁₈ calculated in part (c)

Since we have already found the grams and moles of C₈H₁₈ in part (c), we can use those for this part.

Find moles of CO₂ produced using stoichiometry

Use the balanced equation to set up a ratio between moles of CO₂ and moles of C₈H₁₈: $$\frac{16\ \text{moles of CO₂}}{2\ \text{moles of C₈H₁₈}}=\frac{x\ \text{moles of CO₂}}{\text{moles of C₈H₁₈}}$$ Next, solve for x.

Convert moles of CO₂ to grams

We can use the molar mass of CO₂, which is \(1(12.01 \ \mathrm{g/mol}) + 2(16.00 \ \mathrm{g/mol}) = 44.01 \ \mathrm{g/mol}\). Now convert the moles of CO₂ from step 2 to grams of CO₂: $$\text{grams of CO₂}=\text{moles of CO₂}\times44.01\ \mathrm{g/mol}$$

Key concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to produce heat and light, typically resulting in a flame. In the case of octane, a hydrocarbon, combustion results in the production of carbon dioxide and water. The general formula for combustion of a hydrocarbon can be represented as: - Hydrocarbon + Oxygen \( \rightarrow \) Carbon dioxide + Water
For octane, \[2 \mathrm{C}_{8}\mathrm{H}_{18} (l) + 25 \mathrm{O}_{2} (g) \rightarrow 16 \mathrm{CO}_{2} (g) + 18 \mathrm{H}_{2} \mathrm{O} (g)\] The process exemplifies exothermic reactions, meaning they release energy, which is why combustion reactions are commonly used as a source of energy.
Understanding combustion is essential for applications like automobile engines and energy production, where efficient fuel use is vital.

• This reaction is why gasoline is effective for powering vehicles.
• Due to its completeness, this form of combustion minimizes the release of pollutants, like carbon monoxide.

Molar Mass Calculations

Molar mass calculations help us determine the mass of one mole of any given substance, which is critical in stoichiometry. For any compound, the molar mass is the sum of the atomic masses of all atoms in the formula.
In the exercise, octane (\( \mathrm{C}_{8}\mathrm{H}_{18} \)) is considered.
To calculate its molar mass:

• Each carbon atom has an atomic mass of approximately \(12.01 \mathrm{g/mol}\).
• Each hydrogen atom has an atomic mass of approximately \(1.008 \mathrm{g/mol}\).

For octane: \[8 (12.01) + 18 (1.008) = 114.23 \mathrm{g/mol} \] Calculating molar mass allows us to convert grams to moles, providing a bridge to connect mass to molecular quantities. It is a fundamental in determining the amounts of reactants and products in a given chemical reaction.

Mass-to-Mole Conversions

Mass-to-mole conversions are necessary to connect the physical mass of a substance to its quantity in chemical reactions. By using molar mass as a conversion factor, we can switch from grams to moles, making it easier to apply stoichiometry.
For instance, assuming we have a mass of 10.0 grams of octane (\( \mathrm{C}_{8}\mathrm{H}_{18}\)), the conversion to its mole form is as follows: \[\text{moles of } \mathrm{C}_{8}\mathrm{H}_{18} = \frac{10.0 \text{ grams}}{114.23 \text{ } \mathrm{g/mol}} \]The outcome offers the number of moles, a key step to further apply the stoichiometry of the combustion reaction.
This knowledge is practical for scaling reactions for larger or smaller amounts of substances, allowing for precise chemical manufacturing, laboratory experimentation, and resource management.

Chemical Equation Balancing

Chemical equation balancing is an essential skill in stoichiometry; it involves ensuring both sides of a reaction have the same number of each type of atom, reflecting the conservation of mass. In the combustion of octane, balancing the equation highlights the importance of stoichiometric coefficients. In the provided reaction: \[2 \mathrm{C}_{8}\mathrm{H}_{18}(l) + 25 \mathrm{O}_{2}(g) \rightarrow 16 \mathrm{CO}_{2}(g) + 18 \mathrm{H}_{2} \mathrm{O}(g)\]

• We see that each element (C, H, and O) has the same number of atoms on both sides of the equation.
• This balance allows us to calculate how much of each reactant is needed and how much product will be generated.

Balancing equations ensures no atoms are lost along the way. It ensures that chemical reactions adhere to the physical laws governing them, a crucial step before performing any real-world or laboratory calculations.