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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?

Step by step solution

01

(a) Moles of O鈧� needed for 1.50 mol of C鈧圚鈧佲倛)

For this part, we will use stoichiometry to relate moles of C鈧圚鈧佲倛 and O鈧� 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 O鈧� needed to burn 1.50 moles of C鈧圚鈧佲倛, set up a stoichiometric ratio between moles of O鈧� and moles of C鈧圚鈧佲倛: $$\frac{25\ \text{moles of O鈧倉}{2\ \text{moles of C鈧圚鈧佲倛}}=\frac{x\ \text{moles of O鈧倉}{1.50\ \text{moles of C鈧圚鈧佲倛}}$$ Next, solve for x. (b) Grams of O鈧� needed to burn 10.0 g of C鈧圚鈧佲倛

02

Find moles of C鈧圚鈧佲倛 from given mass

We can use the molar mass of C鈧圚鈧佲倛, 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 C鈧圚鈧佲倛 to moles: $$\text{moles of C鈧圚鈧佲倛}=\frac{10.0\ \mathrm{g}}{114.23\ \mathrm{g/mol}}$$

03

Find moles of O鈧� needed using stoichiometry

Use the balanced equation and the ratio from part (a) to find the moles of O鈧� needed for the moles of C鈧圚鈧佲倛 found in step 1.

04

Convert moles of O鈧� to grams

We can use the molar mass of O鈧�, which is \(2(16.00 \ \mathrm{g/mol}) = 32.00 \ \mathrm{g/mol}\). Now convert the moles of O鈧� from step 2 to grams of O鈧�: $$\text{grams of O鈧倉=\text{moles of O鈧倉\times32.00\ \mathrm{g/mol}$$ (c) Grams of O鈧� required to burn 15.0 gal of C鈧圚鈧佲倛

05

Convert gallons to grams of C鈧圚鈧佲倛

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

06

Find moles of C鈧圚鈧佲倛 from grams

Use the molar mass of C鈧圚鈧佲倛 found in part (b) to convert the grams of C鈧圚鈧佲倛 to moles.

07

Find moles of O鈧� needed using stoichiometry

Use the balanced equation and the ratio from part (a) to find the moles of O鈧� needed for the moles of C鈧圚鈧佲倛 found in step 2.

08

Convert moles of O鈧� to grams

Use the molar mass of O鈧� found in part (b) to convert the moles of O鈧� from step 3 to grams of O鈧�. (d) Grams of CO鈧� produced when 15.0 gal of C鈧圚鈧佲倛 are combusted

09

Use grams or moles of C鈧圚鈧佲倛 calculated in part (c)

Since we have already found the grams and moles of C鈧圚鈧佲倛 in part (c), we can use those for this part.

10

Find moles of CO鈧� produced using stoichiometry

Use the balanced equation to set up a ratio between moles of CO鈧� and moles of C鈧圚鈧佲倛: $$\frac{16\ \text{moles of CO鈧倉}{2\ \text{moles of C鈧圚鈧佲倛}}=\frac{x\ \text{moles of CO鈧倉}{\text{moles of C鈧圚鈧佲倛}}$$ Next, solve for x.

11

Convert moles of CO鈧� to grams

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.