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Chapter 3: Problem 107

(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

Expert verified
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

01

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)
02

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
03

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
04

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
05

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
06

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.
07

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
08

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
09

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
10

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
11

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
12

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.

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Most popular questions from this chapter

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol \((\mathrm{THC}),\) the active ingredient in marijuana, is required to produce intoxication. The molecular formula of \(\mathrm{THC}\) is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). How many moles of THC does this 25 \mug represent? How many molecules? (b) Caffeine, a stimulant found in coffee, contains \(49.5 \% \mathrm{C}\), \(5.15 \% \mathrm{H}, 28.9 \% \mathrm{~N},\) and \(16.5 \% \mathrm{O}\) by mass and has a molar mass of \(195 \mathrm{~g} / \mathrm{mol}\). (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains \(35.51 \% \mathrm{C}, 4.77 \% \mathrm{H}, 37.85 \% \mathrm{O},\) \(8.29 \% \mathrm{~N},\) and \(13.60 \% \mathrm{Na},\) and has a molar mass of \(169 \mathrm{~g} / \mathrm{mol}\)

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