/*! 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 40 \(\Theta\) In the "Chemistry in ... [FREE SOLUTION] | 91影视

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\(\Theta\) In the "Chemistry in Focus" segment Cars of the Future, the claim is made that the combustion of gasoline for some cars causes about 11 b of \(\mathrm{CO}_{2}\) to be produced for each mile traveled. Fstimate the gas mileage of a car that produces about \(11 \mathrm{~b}\) of \(\mathrm{CO}_{2}\) per mile traveled. Assume gasoline has a density of \(0.75 \mathrm{~g} / \mathrm{mL}\) and is \(100 \%\) octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right) .\) While this last part is not true, it is close enough for an estimation. The reaction can be represented by the following unbalanced chemical equation: $$ \mathrm{C}_{8} \mathrm{H}_{18}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} $$

Short Answer

Expert verified
The estimated gas mileage of the car that produces about 11 b of CO鈧 per mile traveled is approximately 7.95 miles per gallon.

Step by step solution

01

Balance the chemical equation

To balance the chemical equation, we need to find the correct coefficients for each chemical species involved. The balanced equation is: \[ \mathrm{C}_{8} \mathrm{H}_{18}+12.5\mathrm{O}_{2} \rightarrow 8\mathrm{CO}_{2}+9\mathrm{H}_{2} \mathrm{O} \]
02

Calculate the molar mass of octane (饾惗_8饾惢_18)

To determine the molar mass, we need to sum the atomic masses of all the atoms in the chemical formula: Molar mass of 饾惗_8饾惢_18 = (8 脳 (12.01 g/mol)) + (18 脳 (1.01 g/mol)) = 96.08 g/mol + 18.18 g/mol = 114.26 g/mol
03

Calculate the mass of CO鈧 produced

We know that 11 b (bornes) of CO鈧 is produced per mile traveled. We need to get this mass in grams. Since 1 b is equivalent to 100 grams, 11 b of CO鈧 is equal to 11脳100 g = 1100 g. Now we can determine the moles of CO鈧 produced using the molar mass of CO鈧: Moles of CO鈧 produced = 1100 g / (44.01 g/mol) = 24.99 mol From the balanced equation, 1 mol of C_8H_18 produces 8 mol of CO鈧. So we can determine the moles of C_8H_18 consumed: Moles of C_8H_18 = moles of CO鈧 produced / 8 = 24.99 mol / 8 = 3.124 mol Now we can calculate the mass of C_8H_18 consumed using its molar mass: Mass of C_8H_18 = moles of C_8H_18 脳 molar mass of C_8H_18 = 3.124 mol 脳 114.26 g/mol = 356.9 g
04

Estimate the gas mileage

We have found the mass of the gasoline (C_8H_18) consumed for one mile traveled (356.9 g). We also know the density of gasoline is 0.75 g/mL. Now, we will find the volume of gasoline consumed: Volume of gasoline = mass of gasoline / density = 356.9 g / (0.75 g/mL) = 475.9 mL Since we have found out the amount of gasoline consumed per mile (475.9 mL), now we can express it as gas mileage: Gas Mileage = 1 mile / volume of gasoline = 1 mile / 475.9 mL 鈮 0.0021 miles/mL To convert to miles per gallon (mpg), we need to remember there are 3,785.41 mL in a gallon: Gas Mileage (mpg) = 0.0021 miles/mL 脳 3,785.41 mL/gallon 鈮 7.95 mpg So, the estimated gas mileage of the car is approximately 7.95 miles per gallon.

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Key Concepts

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

Stoichiometry
Stoichiometry is the area of chemistry that involves calculating the quantities, often in moles, of reactants and products in a chemical reaction. This process involves using balanced chemical equations to ensure the law of conservation of mass is upheld鈥攎eaning atoms are neither created nor destroyed during a reaction.

For students tackling stoichiometric problems, a helpful tip is to always start with a balanced chemical equation. This establishes the correct molar ratios between the different chemical species involved. As showcased in the gasoline combustion example, once the equation is balanced, it's possible to determine how many moles of a product (such as carbon dioxide, CO2) are produced from a specific amount of reactants (like octane, C8H18).

For clarity and to reduce possible errors, students should:
  • Write down the balanced chemical equation
  • Work systematically, calculating one substance at a time
  • Confirm units are consistent (convert if necessary)
  • Double-check molar ratios to avoid miscalculating proportions
Molar Mass Calculation
Understanding how to calculate molar mass is crucial for stoichiometry, as it allows conversion between mass in grams and amount in moles. The molar mass is determined by summing up the atomic masses of every atom in a compound鈥攖aken from the periodic table鈥攎ultiplied by the number of each type of atom present in the molecule.

In solving problems like the one provided鈥攅stimating gas mileage based on CO2 production from gasoline combustion鈥攁ccurately calculating the molar mass of octane, C8H18, is key. This involves multiplying the atomic mass of carbon (12.01 g/mol) by 8, and of hydrogen (1.01 g/mol) by 18, then summing these products.

Students often benefit from breaking this process into steps:

Identify

  • Identify the molecular formula of the substance. In the example, C8H18 for octane.

Calculate

  • Calculate each element's contribution separately, then add them together for total molar mass.

Confirm

  • Always cross-check the calculated molar mass with reliable sources or the periodic table for accuracy.
Gasoline Combustion
Gasoline combustion is a chemical reaction where gasoline reacts with oxygen to produce water (H2O), carbon dioxide (CO2), and energy. This type of reaction is highly exothermic, powering the engines of many vehicles. By approximating gasoline as octane (C8H18), the balanced equation for its combustion in oxygen shows the stoichiometric ratios of the reactants and products.

In the context of the provided exercise, the stoichiometry and balanced equation led to an understanding of the environmental impact鈥攊ndicating how much CO2 is produced鈥攁nd economic aspects, reflecting gas mileage. To improve the estimation of gas mileage, consider:
  • Including the energy content per mole of gasoline, which can show the efficiency of the vehicle's engine
  • Examining different environmental conditions or operating scenarios, which might alter the combustion efficiency

The practical implications of gasoline combustion in automotive engineering and environmental science are significant and multifaceted. Engaging with exercises that involve gasoline combustion allows students to better grasp the importance of such combustion reactions in our daily lives as well as the necessity to accurately understand and calculate their outcomes.

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Most popular questions from this chapter

Barium chloride solutions are used in chemical analysis for the quantitative precipitation of sulfate ion from solution. $$ \mathrm{Ba}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightarrow \mathrm{BaSO}_{4}(s) $$ Suppose a solution is known to contain on the order of \(150 \mathrm{mg}\) of sulfate ion. What mass of barium chloride should be added to guarantee precipitation of all the sulfate ion?

Consider the following unbalanced chemical equation. $$ \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(67.4 \mathrm{~g}\) of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

According to his prelaboratory theoretical yield calculations, a student's experiment should have produced \(1.44 \mathrm{~g}\) of magnesium oxide. When he weighed his product after reaction, only \(1.23 \mathrm{~g}\) of magnesium oxide was present. What is the student's percent yicld?

Using the average atomic masses given inside the front cover of the text, calculate how many moles of each substance the following masses represent. a. 4.21 g of copper(II) sulfate b. \(7.94 \mathrm{~g}\) of barium nitrate c. \(1.24 \mathrm{mg}\) of water d. \(9.79 \mathrm{~g}\) of tungsten c. 1.45 lb of sulfur f. 4.65 g of ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) g. 12.01 g of carbon

For each of the following tunbalanced equations, calculate how many moles of the second reactant would be required to react completely with 0.557 grams of the first reactant. a. \(\mathrm{Al}(s)+\mathrm{Br}_{2}(l) \rightarrow \mathrm{AlBr}_{3}(s)\) b. \(\mathrm{Hg}(s)+\mathrm{HClO}_{4}(a q) \rightarrow \mathrm{Hg}\left(\mathrm{ClO}_{4}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{K}(s)+\mathrm{P}(s) \rightarrow \mathrm{K}_{3} \mathrm{P}(s)\) d. \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{CCl}_{4}(l)+\mathrm{HCl}(g)\)

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