Problem 97 - Chemistry The Molecular Science

Question

Assume that a car burns pure octane, $$ \mathrm{C}_{8} \mathrm{H}_{18}\left(d=0.703 \mathrm{~g} / \mathrm{cm}^{3}\right) $$ (a) Write the balanced equation for burning octane in air, forming \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (b) If the car has a fuel efficiency of 32 miles per gallon of octane, what volume of \(\mathrm{CO}_{2}\) at \(25^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\) is generated when the car goes on a 10.0 -mile trip? (The volume of \(1 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is \(\left.24.5 \mathrm{~L} .\right)\)

Step-by-Step Solution

Answer

The car produces 1426.9 liters of \(\mathrm{CO}_2\) during the 10-mile trip.

Step 1: Write the Balanced Equation

The balanced chemical equation for the combustion of octane \( \mathrm{C}_8\mathrm{H}_{18} \) is obtained by balancing carbons, hydrogens, and oxygens. The equation is: \[ \mathrm{2C}_8\mathrm{H}_{18} + 25\mathrm{O}_2 \rightarrow 16\mathrm{CO}_2 + 18\mathrm{H}_2\mathrm{O} \]

Step 2: Determine Volume of Octane Consumed

First, calculate how much octane is consumed in the 10-mile trip. The fuel efficiency is 32 miles per gallon, so the amount of octane used is \( \frac{10 \text{ miles}}{32 \text{ miles/gallon}} = 0.3125 \text{ gallons} \). Convert this to liters using \( 1 \text{ gallon} = 3.78541 \text{ liters} \), which gives us \( 0.3125 \times 3.78541 = 1.182 \text{ liters} \).

Step 3: Convert Volume to Mass of Octane

Use the density of octane to find the mass: \( \text{Mass} = \text{Volume} \times \text{Density} = 1.182 \text{ liters} \times 0.703 \text{ g/cm}^3 = 831.7 \text{ grams} \). Convert this mass from grams to moles using the molar mass of octane \( \mathrm{C}_8\mathrm{H}_{18} \) which is \( 114.22 \text{ g/mol} \): \( \text{Moles of octane} = \frac{831.7 \text{ grams}}{114.22 \text{ g/mol}} = 7.28 \text{ moles} \).

Step 4: Calculate Moles of \(\mathrm{CO}_2\) Produced

According to the balanced equation, burning 2 moles of \( \mathrm{C}_8\mathrm{H}_{18} \) yields 16 moles of \( \mathrm{CO}_2 \). Therefore, \( \frac{16}{2} = 8 \text{ moles of } \mathrm{CO}_2 \) are produced per mole of octane burned. For 7.28 moles of octane, \( 8 \times 7.28 = 58.24 \text{ moles of } \mathrm{CO}_2 \) are produced.

Step 5: Calculate Volume of \(\mathrm{CO}_2\) Produced

Use the ideal gas law to find the volume of \( \mathrm{CO}_2 \) produced. The volume of 1 mole \( \mathrm{CO}_2 \) is \( 24.5 \text{ liters} \) at the given conditions. Thus: \( 58.24 \text{ moles} \times 24.5 \text{ L/mol} = 1426.9 \text{ liters of } \mathrm{CO}_2 \).

Key Concepts

Balanced Chemical Equation

In any combustion reaction, balancing the chemical equation is crucial. This ensures the law of conservation of mass is followed, meaning matter cannot be created or destroyed. For the combustion of octane (\( \mathrm{C}_8\mathrm{H}_{18} \)), we need to balance the number of carbon, hydrogen, and oxygen atoms on both sides of the equation.

The balanced equation for the combustion of octane in air, producing carbon dioxide (\( \mathrm{CO}_2 \)) and water (\( \mathrm{H}_2\mathrm{O} \)), is:\[ \mathrm{2C}_8\mathrm{H}_{18} + 25\mathrm{O}_2 \rightarrow 16\mathrm{CO}_2 + 18\mathrm{H}_2\mathrm{O} \]

• The equation shows that 2 molecules of octane react with 25 molecules of oxygen to form 16 molecules of carbon dioxide and 18 molecules of water.
• Start by balancing the carbon atoms first, followed by hydrogen, and finally balance the oxygen.
• Remember, coefficients in front of each compound reflect the stoichiometric ratios, necessary to ensure equal numbers of atoms of each element.

Ideal Gas Law

The ideal gas law connects the physical quantities of gases, allowing us to predict volumes under different conditions. The law is represented by the equation: \[ PV = nRT \]where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the universal gas constant (\( 0.0821 \text{ L atm/mol K} \))
• \( T \) is the temperature in Kelvin

For our problem, we are given the conditions of the volume of one mole of carbon dioxide as 24.5 liters under specific conditions (25°C and 1 atm).

Using the ideal gas law allows us to calculate the volume of CO2 produced in the reaction by determining the number of moles and multiplying by 24.5 L/mol.

Fuel Efficiency

Fuel efficiency indicates how effectively a vehicle uses fuel to travel, often expressed as miles per gallon (mpg). A higher mpg value suggests that the vehicle consumes less fuel over a fixed distance, which is a measure of performance and economy.

In this problem, a car with a fuel efficiency of 32 mpg needs to be analyzed. For a trip length of 10 miles, we calculate the fuel consumption:

• First, divide the distance by the efficiency: \( \frac{10 \text{ miles}}{32 \text{ miles/gallon}} = 0.3125 \text{ gallons} \)
• Then, convert gallons to liters using the conversion factor: 1 gallon = 3.78541 liters, which gives us 1.182 liters of octane used.

Understanding fuel efficiency not only helps in determining the fuel consumption but also plays a significant role in calculating emission outputs.

Stoichiometry

Stoichiometry is about the quantitative aspects of chemical reactions, focusing on the relationships between reactants and products in a balanced equation.

With stoichiometry, you determine how much of a product will form from known amounts of reactants:

• The balanced equation shows that each 2 moles of \( \mathrm{C}_8\mathrm{H}_{18} \) produces 16 moles of\( \mathrm{CO}_2 \).
• This gives a stoichiometric ratio of 8:1 for \( \mathrm{CO}_2:moles \) of \( \mathrm{C}_8\mathrm{H}_{18} \).

In application:

• From 1.182 liters of octane, the moles of octane are calculated as 7.28 moles using its molar mass.
• From this, the moles of \( \mathrm{CO}_2 \) produced are 58.24 by multiplying the moles of octane by the stoichiometric ratio.
• The volume of \( \mathrm{CO}_2 \) is then found using the ideal gas law principles to calculate 1,426.9 liters of carbon dioxide output from the reaction.