/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 Driving 1000 miles a month is no... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 68

Driving 1000 miles a month is not unusual for a short-distance commuter. If your vehicle gets 25 mpg, you would use 40 gallons \((\approx 150 \mathrm{L})\) of gasoline every month. If gasoline is approximated as \(\mathrm{C}_{8} \mathrm{H}_{18}(d=0.703 \mathrm{g} / \mathrm{mL}),\) how much carbon dioxide does your vehicle emit every month? The unbalanced chemical equation for the reaction is$$ \mathrm{C}_{8} \mathrm{H}_{18}(\ell)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$.

Short Answer

Expert verified
Answer: The vehicle emits 327,967.38 grams of carbon dioxide per month.

Step by step solution

01

Calculate the volume of gasoline used per month

First, we need to calculate the volume of gasoline the vehicle uses per month. We are given that the vehicle drives 1000 miles a month and gets 25 miles per gallon (mpg). We can use the following formula: Volume of gasoline used = Total miles driven / miles per gallon In this case, Volume of gasoline used = 1000 miles / 25 mpg = 40 gallons
02

Convert volume of gasoline used to mass

Next, we need to convert the volume of gasoline used to mass. To do this, we will first convert the volume from gallons to milliliters and then use the density of gasoline (0.703 g/mL) to calculate the mass. 1 gallon = 3,785.41 mL 40 gallons = 40 脳 3,785.41 mL = 151,416.4 mL Now, we can use the density of gasoline to find the mass: Mass of gasoline used = Volume 脳 Density Mass of gasoline used = 151,416.4 mL 脳 0.703 g/mL = 106,494.34 g
03

Calculate the moles of C8H18 consumed

Now, we need to determine the number of moles of C8H18 consumed. To do this, we will use the molar mass of C8H18 (114.23 g/mol): Moles of C8H18 = Mass of gasoline used / Molar mass Moles of C8H18 = 106,494.34 g / 114.23 g/mol = 932.32 mol
04

Balance the chemical equation

In this step, we will balance the chemical equation for the reaction between octane (C8H18) and oxygen (O2): C8H18(l) + O2(g) 鈫� CO2(g) + H2O(g) The balanced equation is: 2 C8H18(l) + 25 O2(g) 鈫� 16 CO2(g) + 18 H2O(g)
05

Calculate the moles of CO2 produced

Now that the equation is balanced, we can use the stoichiometry to find the moles of CO2 produced: For every 2 moles of C8H18, 16 moles of CO2 are produced. Therefore, we need to find the amount of CO2 produced for the number of moles of C8H18 consumed: Moles of CO2 = (Moles of C8H18 脳 Moles of CO2 produced) / Moles of C8H18 consumed Moles of CO2 = (932.32 mol 脳 16) / 2 = 7,458.56 mol
06

Calculate the mass of CO2 emitted

Finally, we need to calculate the mass of CO2 emitted per month using the molar mass of CO2 (44.01 g/mol): Mass of CO2 emitted = Moles of CO2 脳 Molar mass Mass of CO2 emitted = 7,458.56 mol 脳 44.01 g/mol = 327,967.38 g So, the vehicle emits 327,967.38 grams of carbon dioxide per month.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Chemical Reactions
Chemical reactions are processes where reactants transform into products. These transformations follow the law of conservation of matter, which states that matter cannot be created or destroyed in an isolated system. During the combustion of gasoline, a type of chemical reaction, each carbon (C) atom from the gasoline reacts with oxygen (O2) to produce carbon dioxide (CO2). Similarly, hydrogen (H) atoms combine with oxygen to form water (H2O). Balancing the chemical equation is indeed essential as it ensures that the number of atoms for each element is the same on both sides of the reaction, fulfilling the conservation of matter.

For example, in the balanced equation for the combustion of octane (C8H18), a component of gasoline, it shows that 2 moles of octane react with 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O. This step helps in calculating the correct amount of CO2 produced as a result of combustion.
Molar Mass Calculation Essentials
Molar mass calculation is a fundamental concept in chemistry that involves determining the mass of one mole of a substance. The molar mass is expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all the atoms in a molecule. For stoichiometry problems, like the combustion of gasoline, understanding molar mass is crucial.

When calculating the molar mass of octane (C8H18), we add the masses of 8 carbon atoms and 18 hydrogen atoms. Once we have the molar mass, we can convert between mass and moles of a substance, which allows us to use stoichiometry to predict the quantities of reactants needed and products formed in a chemical reaction. For instance, knowing the molar mass of octane and carbon dioxide is essential to determine how much CO2 will be emitted when a certain amount of octane is burned.
The Science Behind Gasoline Combustion
Gasoline combustion is an exothermic chemical reaction that releases energy by breaking down complex hydrocarbons into simpler molecules, primarily CO2 and water vapor. This process is used in internal combustion engines to power vehicles. The density of gasoline (0.703 g/mL) is a critical factor because it allows us to convert between volume and mass 鈥� a necessary step in solving stoichiometry problems.

In the complete combustion of gasoline, represented simplistically by octane (C8H18), a precise quantity of oxygen is required for every molecule of gasoline. The reaction produces a specific quantity of CO2, determined by the stoichiometric coefficients in the balanced chemical equation. Accurate molar mass calculation and understanding of stoichiometry are imperative for predicting the mass of CO2 emitted from burning a given amount of gasoline, which is a common concern for environmental impact assessments.

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Most popular questions from this chapter

Surgical-Grade Titanium Medical implants and high-quality jewelry items for body piercings are frequently made of a material known as G23Ti, or surgical- grade titanium. The percent composition of the material is \(64.39 \%\) titanium,\(24.19 \%\) aluminum, and \(11.42 \%\) vanadium. What is the empirical formula for surgical-grade titanium?

Some indoor air-purification systems work by converting a little of the oxygen in the air to ozone, which oxidizes mold and mildew spores and other biological air pollutants. The chemical equation for the ozone generation reaction is,$$3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{O}_{3}(g)$$.It is claimed that one such system generates \(4.0 \mathrm{g}\) of \(\mathrm{O}_{3}\) per hour from dry air passing through the purifier at a flow of \(5.0 \mathrm{L} / \mathrm{min.}\) If 1 liter of indoor air contains \(0.28 \mathrm{g}\) of \(\mathrm{O}_{2}\). a. what fraction of the molecules of \(\mathrm{O}_{2}\) is converted to \(\mathrm{O}_{3}\) by the air purifier? b. what is the percent yield of the ozone generation reaction?

A 1 -liter sample of seawater contains \(19.4 \mathrm{g}\) of \(\mathrm{Cl}^{-}, 10.8 \mathrm{g}\) of \(\mathrm{Na}^{+},\) and \(1.29 \mathrm{g}\) of \(\mathrm{Mg}^{2+}\) a. How many moles of each ion are present? b. If we evaporated the seawater, would there be enough Cl "present to form the chloride salts of all the sodium and magnesium present?

Silane \(\left(\mathrm{SiH}_{4}\right)\) is used in the electronics industry to manufacture thin films of silicon. What is the percent Si by mass in \(\mathrm{SiH}_{4} ?\) Does silane have the same percent Si by mass as disilane, \(\mathrm{Si}_{2} \mathrm{H}_{6} ?\)

Large quantities of fertilizer are washed into the Mississippi River from agricultural land in the Midwest. The excess nutrients collect in the Gulf of Mexico, promoting the growth of algae and endangering other aquatic life.a. One commonly used fertilizer is ammonium nitrate. What is the chemical formula of ammonium nitrate? b. Corn farmers typically use \(5.0 \times 10^{3} \mathrm{kg}\) of ammonium nitrate per square kilometer of cornfield per year. Ammonium nitrate can be prepared by the following reaction: $$\mathrm{NH}_{3}(a q)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{NH}_{4} \mathrm{NO}_{3}(a q)$$ How much nitric acid would be required to make the fertilizer needed for \(1 \mathrm{km}^{2}\) of cornfield per year? c. The ammonium ions can be converted into \(\mathrm{NO}_{3}^{-}\) by bacterial action. \(\mathrm{NH}_{4}^{+}(a q)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{NO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{H}^{+}(a q)\) If \(10 \%\) of the ammonium component of \(5.0 \times 10^{2} \mathrm{kg}\) of fertilizer ends up as nitrate, how much oxygen would be consumed?
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