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Chapter 3: Problem 110

(a) If an automobile travels \(350 \mathrm{~km}\) with a gas mileage of 9.0 \(\mathrm{km} / \mathrm{L}\), how many kilograms of \(\mathrm{CO}_{2}\) are produced? Assume that the gasoline is composed of octane, $\mathrm{C}_{8} \mathrm{H}_{18}(l),\( whose density is \)0.692 \mathrm{~g} / \mathrm{mL}$. (b) Repeat the calculation for a truck that has a gas mileage of $2 \mathrm{~km} / \mathrm{L}$.

Short Answer

Expert verified
The automobile produces approximately 84.4 kg of CO₂, while the truck produces approximately 378 kg of CO₂.

Step by step solution

01

Find the amount of gasoline consumed by the automobile

Divide the distance driven (350 km) by the gas mileage (9 km/L) to determine the amount of gasoline consumed: Gasoline_consumed = 350 km / 9.0 km/L = 38.89 L (approx.)
02

Calculate the mass of gasoline consumed by the automobile

Given the density of octane as 0.692 g/mL, we can convert the amount of gasoline consumed to mass: Gas_mass_car = Gasoline_consumed * 0.692 g/mL * 1000 mL/L = 26900 g (approx.)
03

Determine the stoichiometry of the combustion reaction of octane

The balanced equation for the combustion of octane is: C8H18(l) + 12.5 O2(g) -> 8 CO2(g) + 9 H2O(l) For each mole of octane combusted, 8 moles of CO2 are produced.
04

Calculate the molar mass of octane and CO2

Molar mass of octane (C8H18) = (8 * 12.01 g/mol) + (18 * 1.01 g/mol) = 114.23 g/mol Molar mass of CO2 = (1 * 12.01 g/mol) + (2 * 16.00 g/mol) = 44.01 g/mol
05

Calculate the mass of CO2 produced by the automobile

Use stoichiometry to find the mass of CO2 produced from the mass of octane burnt: Mass_CO2_car = (Gas_mass_car /114.23 g/mol) * 8 mol CO2/1 mol C8H18 * 44.01 g/mol = 84400 g or 84.4 kg of CO2 (approx.) b) Calculating the amount of CO2 produced by the truck
06

Find the amount of gasoline consumed by the truck

Divide the distance driven (350 km) by the gas mileage (2 km/L) to determine the amount of gasoline consumed by the truck. Gasoline_consumed_truck = 350 km / 2.0 km/L = 175 L
07

Calculate the mass of gasoline consumed by the truck

Using the density of octane (0.692 g/mL), we can convert the amount of gasoline consumed by the truck: Gas_mass_truck = Gasoline_consumed_truck * 0.692 g/mL * 1000 mL/L = 121000 g (approx.)
08

Use the stoichiometry of the combustion reaction to determine the mass of CO2 produced by the truck

Using the stoichiometry from the previous part, we calculate the mass of CO2 produced by the truck: Mass_CO2_truck = (Gas_mass_truck /114.23 g/mol) * 8 mol CO2/1 mol C8H18 * 44.01 g/mol = 378000 g or 378 kg of CO2 (approx.) The automobile produces approximately 84.4 kg of CO2, while the truck produces approximately 378 kg of CO2.

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When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right),\) bromobenzene $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)$ is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{~g}\), what is the percentage yield?

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