Problem 68 - Chemistry The Central Science In Si Units

Question

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) a 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 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

Answer

(a) 18.75 mol of O2. (b) 35.04 g of O2. (c) 137588.0 g of O2. (d) 121060.51 g of CO2.

Step 1: Balanced Chemical Equation

First, note the balanced chemical equation for the combustion of octane: \[ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l) + 25 \mathrm{O}_{2}(g) \rightarrow 16 \mathrm{CO}_{2}(g) + 18 \mathrm{H}_{2}\mathrm{O}(g) \] This equation indicates the stoichiometric ratio for the reaction.

Step 2: Mole Calculation for O2 (Part a)

To find how many moles of \( \mathrm{O}_{2} \) are needed to burn \( 1.50 \) mol of \( \mathrm{C}_{8} \mathrm{H}_{18} \), use the stoichiometry from the balanced equation. - According to the equation, 2 moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \) require 25 moles of \( \mathrm{O}_{2} \). - Therefore, \( 1.50 \text{ mol of } \mathrm{C}_{8} \mathrm{H}_{18} \) will require \( 1.50 \times \frac{25}{2} = 18.75 \) moles of \( \mathrm{O}_{2} \).

Step 3: Mass Calculation for O2 (Part b)

First, find the moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \) from 10.0 g. - Molar mass of \( \mathrm{C}_{8} \mathrm{H}_{18} \) is \( 114.22 \text{ g/mol} \). - Moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \) = \( \frac{10.0}{114.22} \approx 0.0876 \text{ mol} \) Next, use stoichiometry:- \( 0.0876 \text{ mol of } \mathrm{C}_{8} \mathrm{H}_{18} \) will need \( 0.0876 \times \frac{25}{2} = 1.095 \) moles of \( \mathrm{O}_{2} \). Finally, convert moles of \( \mathrm{O}_{2} \) to grams. - Molar mass of \( \mathrm{O}_{2} \) = \( 32.00 \text{ g/mol} \). - Mass of \( \mathrm{O}_{2} \) = \( 1.095 \times 32.00 = 35.04 \text{ g} \).

Step 4: Volume and Mass of Octane to O2 Mass (Part c)

First, calculate the mass of \( 15.0 \text{ gal} \) of octane: - Convert gallons to liters: \( 15.0 \times 3.78541 = 56.78115 \text{ L} \). - Convert liters to milliliters: \( 56.78115 \text{ L} = 56781.15 \text{ mL} \). - Use density to find mass: \( 56781.15 \times 0.692 \approx 39288.4 \text{ g} \). Calculate moles of octane: - Moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \) = \( \frac{39288.4}{114.22} \approx 343.97 \text{ mol} \). Calculate mass of \( \mathrm{O}_{2} \) needed: - Moles of \( \mathrm{O}_{2} \) required = \( 343.97 \times \frac{25}{2} = 4299.625 \text{ mol} \). - Mass of \( \mathrm{O}_{2} \) = \( 4299.625 \times 32.00 \approx 137588.0 \text{ g} \).

Step 5: Producing CO2 (Part d)

Starting with mass of octane as calculated previously, \( 39288.4 \text{ g} \): Using moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \): - Moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \) = \( 343.97 \text{ mol} \). Use stoichiometry to find moles of \( \mathrm{CO}_{2} \): - From the balanced equation, 2 moles of \( \mathrm{C}_{8} \mathrm{H}_{18} \) produce 16 moles of \( \mathrm{CO}_{2} \). - \( 343.97 \text{ mol} \times \frac{16}{2} = 2751.76 \text{ mol of } \mathrm{CO}_{2} \). Convert moles of \( \mathrm{CO}_{2} \) to grams: - Molar mass of \( \mathrm{CO}_{2} \) = \( 44.01 \text{ g/mol} \). - Mass of \( \mathrm{CO}_{2} \) = \( 2751.76 \times 44.01 \approx 121060.51 \text{ g} \).

Key Concepts

Stoichiometry: Understanding Chemical Quantities

Stoichiometry is a core concept that helps us understand the relationships in a chemical reaction. It is based on the idea of conserving mass during a reaction.
The balanced chemical equation for the combustion of octane gives us the stoichiometric ratios:

• The equation shows that 2 moles of octane react with 25 moles of oxygen to produce combustion products.
• This ratio is vital for calculations; it tells us how much of one reactant is needed to react with a certain amount of another substance.

When dealing with real-world chemical problems, these stoichiometric coefficients guide conversions between moles of different substances.
This method allows us to calculate the exact quantities of required materials or produced substances in a reaction by simply using proportions from the balanced chemical equation.

Molar Mass: Translating Moles into Grams

Molar mass is the weight of one mole of any given substance and is expressed in grams per mole (g/mol). This property allows us to convert between grams and moles – a crucial step in many chemistry problems.

• For octane (\( \mathrm{C}_{8} \mathrm{H}_{18} \)), the molar mass is calculated by adding the atomic masses of carbon (\( \approx 12.01 \text{ g} \)) and hydrogen (\( \approx 1.01 \text{ g} \)) in the molecule.
• The molar mass of octane turns out to be \( 114.22 \text{ g/mol} \).

When considering oxygen gas (\( \mathrm{O}_{2} \)), the molar mass is simply twice the atomic mass of an oxygen atom, resulting in \( 32.00 \text{ g/mol} \).
Determining the molar mass of each component allows us to convert amounts expressed in grams to moles and vice versa, thereby facilitating various computations in reaction equations.

Gas Volume Conversion: Translations from Mole to Volume

In chemistry, converting between the volume of gases and their quantities in moles is often necessary, especially under specific conditions of temperature and pressure. This is particularly relevant for reactions involving gases like the combustion of octane.

• The Ideal Gas Law, expressed as \( PV = nRT \), is often used for such conversions, where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the gas constant, and \( T \) is temperature.
• At standard temperature and pressure (STP: 0°C, 1 atm), 1 mole of any gas occupies 22.414 liters.

These conversions are crucial when dealing with changes in physical conditions or calculating reactant or product volumes, ensuring we consider the physical state of each component throughout the reaction.

Density Calculation: From Volume to Mass

Density is defined as the mass of a substance per unit volume, typically expressed as grams per milliliter (g/mL). This characteristic helps convert between volume and mass, especially for liquids like octane.

• In the combustion example, density is used to find the mass from a given volume of octane.
• With a density of \( 0.692 \text{ g/mL} \), converting 15.0 gallons of octane to liters (since 1 gallon = 3.78541 liters), and then to milliliters, allows calculation of the total mass.

By connecting volume and mass through density, we can easily transition between these quantities, ensuring accurate scaling for different uses in both laboratory and industrial settings.