Question

(a) If an automobile travels \(225 \mathrm{mi}\) with a gas mileage of \(20.5 \mathrm{mi} / \mathrm{gal}\), how many kilograms of \(\mathrm{CO}_{2}\) are produced? Assume that the gasoline is composed of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) whose density is \(0.69 \mathrm{~g} / \mathrm{mL}\). (b) Repeat the calculation for a truck that has a gas mileage of \(5 \mathrm{mi} / \mathrm{gal}\).

Short answer

The automobile produces 88.53 kg of CO2, while the truck produces 362.96 kg of CO2 when traveling 225 miles.

Step-by-step solution

Part (a): Miles to Gallons

To convert the miles traveled into gallons consumed, we can use the automobile's given gas mileage. We have: Gas mileage = 20.5 mi/gal Miles traveled = 225 mi Gallons consumed = Miles traveled / Gas mileage = 225 mi / 20.5 mi/gal = 10.98 gal (approximately 11 gal)

Part (a): Gallons to Liters

Next, we can convert gallons to liters using the conversion factor 1 gal = 3.785 L. So, Liters of gasoline = 10.98 gal × 3.785 L/gal = 41.57 L

Part (a): Mass of Gasoline

Now, let's calculate the mass of gasoline using the density of octane: Density of octane = 0.69 g/mL L to mL conversion: 1 L = 1000 mL Mass of gasoline = 41.57 L × 1000 mL/L × 0.69 g/mL = 28684 g

Part (a): Moles of Octane

To find the moles of octane consumed, we'll use the molar mass of octane, which is approximately 114 g/mol. Moles of octane = Mass of gasoline / Molar mass of octane = 28684 g / 114 g/mol = 251.61 mol

Part (a): Moles of CO2 Produced

The balanced chemical equation for the combustion of octane is: C8H18 + 12.5 O2 → 8 CO2 + 9 H2O From the balanced equation, 1 mole of octane produces 8 moles of CO2. Therefore, Moles of CO₂ = Moles of octane × 8 = 251.61 mol × 8 = 2012.9 mol

Part (a): Mass of CO2 Produced

Finally, we can calculate the mass of CO2 produced using the molar mass of CO2, which is approximately 44 g/mol. Mass of CO₂ = Moles of CO₂ × Molar mass of CO₂ = 2012.9 mol × 44 g/mol = 88527 g Now, convert the grams to kilograms: 88527 g × (1 kg / 1000 g) = 88.53 kg So, the automobile produces 88.53 kg of CO2.

Part (b): Truck's Miles to Gallons

For part (b), we need to repeat the calculation for a truck with a gas mileage of 5 mi/gal. First, we have: Gas mileage = 5 mi/gal Miles traveled = 225 mi Gallons consumed = Miles traveled / Gas mileage = 225 mi / 5 mi/gal = 45 gal

Part (b): Gallons to Liters

Next, we can convert gallons to liters again using the conversion factor 1 gal = 3.785 L. So, Liters of gasoline = 45 gal × 3.785 L/gal = 170.33 L

Part (b): Mass of Gasoline

Calculate the mass of gasoline for the truck: Mass of gasoline = 170.33 L × 1000 mL/L × 0.69 g/mL = 117526 g

Part (b): Moles of Octane

Find the moles of octane consumed for the truck: Moles of octane = Mass of gasoline / Molar mass of octane = 117526 g / 114 g/mol = 1031.12 mol

Part (b): Moles of CO2 Produced

Determine the moles of CO2 for the truck: Moles of CO₂ = Moles of octane × 8 = 1031.12 mol × 8 = 8249 mol

Part (b): Mass of CO2 Produced

Calculate the mass of CO2 produced by the truck: Mass of CO₂ = Moles of CO₂ × Molar mass of CO₂ = 8249 mol × 44 g/mol = 362956 g Now, convert the grams to kilograms: 362956 g × (1 kg / 1000 g) = 362.96 kg So, the truck produces 362.96 kg of CO2.

Key concepts

Stoichiometry

Stoichiometry is like the math of chemistry, helping us figure out the amounts of substances that get consumed and produced in chemical reactions. Think of it as a blueprint that guides chemical processes just as a recipe dictates cooking.
When you know the number of molecules participating in a reaction and the balanced chemical equation, you can predict how much of each product will form.
For instance, in our initial problem of octane combustion - the reaction between octane (\(\mathrm{C}_{8}\mathrm{H}_{18}\))and oxygen (\(\mathrm{O}_{2}\)), the stoichiometry tells us that burning 1 mole of octane generates 8 moles of carbon dioxide (\(\mathrm{CO}_{2}\)):

• Chemical Reaction: \(\mathrm{C}_{8}\mathrm{H}_{18} + 12.5\,\mathrm{O}_{2} \rightarrow 8\,\mathrm{CO}_{2} + 9\,\mathrm{H}_{2}\mathrm{O}\)
• This reaction helps us transform calculations from fuel (octane) to emissions (carbon dioxide).

By understanding stoichiometry, you can translate between masses and moles of reactants and products, which is essential in solving real-world chemistry problems!

Gas Mileage Calculation

Gas mileage calculation is about measuring how far a vehicle travels on a specific amount of fuel. It's usually expressed in miles per gallon (\(\text{mi/gal}\))or kilometers per liter (\(\text{km/L}\)).
Knowing this measurement helps us determine how much fuel a vehicle will use for a trip and the emissions it will produce.For example, using the provided calculator for the car in our example:

• Gas mileage is \(20.5\ \text{mi/gal}\).
• Distance driven is \(225\ \text{mi}\).
• The calculation is done as follows: \(\text{Gallons consumed} = \frac{\text{Miles traveled}}{\text{Gas mileage}} = \frac{225\ \text{mi}}{20.5\ \text{mi/gal}}\approx 11\ \text{gal}\).

This kind of calculation also has practical implications, including saving money on fuel and minimizing emissions, making it a critical aspect of assessing and improving eco-friendliness in transportation.

Mass Conversion in Chemical Reactions

Mass conversion in chemical reactions is an integral part of understanding how different substances change during a reaction. It's about transforming between different units of measurement such as liters, grams, moles, etc.
In our octane combustion example, we see several conversions:

• First, converting the volume of gasoline from gallons to liters, as 1 gallon equals 3.785 liters.
• Then, converting the volume in liters to mass in grams using density; in this case, octane has a density of 0.69 g/mL.
• For instance, \(41.57\ \text{L}\times 1000\,\text{mL/L} \times 0.69\,\text{g/mL} = 28684\,\text{g}\), which transforms to moles: \(\frac{28684\,\text{g}}{114\,\text{g/mol}}\approx 251.61\,\text{mol of octane}\).

Converting mass based on chemical equations ensures that we fully grasp the impact of chemistry on the environment and industrial processes, aiding in both scientific study and practical applications.