Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 8, Problem ISP

In a typical automobile engine, a gasoline vapor-air mixture is compressed and ignited in the cylinders of the engine. This results in a combustion reaction that produces mainly carbon dioxide and water vapor. For simplicity, assume that the fuel is C8H18 and has a density of 0.760 g/mL.

  1. (a) Calculate the partial pressures of N2 and 02 in the air before it goes into the cylinder; assume the atmospheric pressure is 734 mmHg.
  2. (b) Consider the case where the air, without any fuel added, is compressed in the cylinder to seven times atmospheric pressure, the compression ratio of many modem automobile engines. Calculate the partial pressures of N2 and O2 at this pressure.
  3. (c) Now consider the case where 0.050 mL gasoline is added to the air in the cylinder just before compression and completely vaporized. Assume that the volume of the cylinder is 485 mL and the temperature is 150°C. Calculate the partial pressure of the gasoline vapor.
  4. (d) Calculate the amount (mol) of oxygen required to bum the gasoline in part (c) completely to CO2 and H2O.
  5. (e) The combustion reaction in the cylinder creates temperatures in excess of 1200K. Due to the high temperature, some of the nitrogen and oxygen in the air reacts to form nitrogen monoxide. If 10% of the nitrogen is converted to NO, calculate the mass (g) of NO produced by this combustion.
  6. (f) Hot-rod cars use another oxide of nitrogen, dinitrogen monoxide, to create an extra burst in power. When such a power boost is needed, dinitrogen monoxide gas is injected into the cylinders where it reacts with oxygen to form NO. Calculate the mass of dinitrogen monoxide that would have to be injected to form the same quantity of NO as produced in part (e). Assume that sufficient oxygen is present to do so.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Partial pressure of N2 and O2 has to be calculated when the atmospheric pressure is 734 mmHg.

Concept Introduction:

Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

Xa=nantotal

Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

  Pa=Xa×Ptotal

Answer to Problem ISP

Partial pressure of N2 is 573mmHg.

Partial pressure of O2 is 154mmHg.

Explanation of Solution

Percentage by volume of N2 is 78.084%.

Percentage by volume of O2 is 20.948%.

Mole fraction of N2 and O2 is calculated as follows,

  XN2=%N2100 %=78.084%100%=0.78084XO2=%O2100 %=20.948%100%=0.20948

Partial pressure of N2 is calculated as follows,

  PN2=XN2Ptotal =(0.78084)(734mmHg)=573mmHg

Partial pressure of O2 is calculated as follows,

  PO2=XO2Ptotal =(0.20948)(734mmHg)=154mmHg

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Partial pressure of N2 and O2 when the volume of gas is compressed by seven times has to be calculated.

Answer to Problem ISP

Partial pressure of N2 is 4012mmHg.

Partial pressure of O2 is 1076mmHg.

Explanation of Solution

At 734mmHg, partial pressure of N2 and O2 is 573mmHg and 154mmHg respectively.

According to Boyle’s law, pressure and volume is inversely proportional to each other at constant temperature and number of molecules.

  P1V

Hence, when the volume of gas is compressed by seven times, then pressure of the sample is increased by seven times.

Therefore,

Partial pressure of N2 and O2 when the pressure is raised by a factor of 7 is calculated as shown below,

  PN2=7×573mmHg=4012mmHgPO2=7×154mmHg=1076mmHg

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Partial pressure of the gasoline vapor has to be calculated.

Concept Introduction:

Ideal gas Equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas. Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained. It is referred as ideal gas equation.

  PV = nRT

Here,

  n is the moles of gas

    P is the Pressure

  V is the Volume

    T is the Temperature

    R is the gas constant

Answer to Problem ISP

Partial pressure of the gasoline vapor is 18 mmHg.

Explanation of Solution

Given information is shown below,

  VC8H18 = 0.050 mLDensity= 0.760 g/mLVcylinder = 485 mL = 0.485 LT = 150°C = (150+273) = 423 K

Number of moles of gasoline (C8H18) can be determined form the density and molar mass C8H18 as shown below,

  n = 0.050 mL C8H18×0.760 g C8H181 mL C8H18×1 mol C8H18114.2 g C8H18=3.3×104molC8H18

Partial pressure of the gasoline vapor can be calculated using Ideal Gas equation as follows,

  PV=nRTP(0.485L)= (3.3×104molC8H18)(0.0821 L.atm.mol1K1)(423 K)=(3.3×104molC8H18)(0.0821 L.atm.mol1K1)(423 K)(0.485 L)=0.024 atmC8H18= 0.024 atm×760 mmHg1 atm= 18 mmHgC8H18

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Number of moles of O2 required to burn C8H18 has to be calculated.

Answer to Problem ISP

Number of moles of O2 reacted is 0.0041 mol.

Explanation of Solution

Balanced equation for the combustion of gasoline is given below,

  2C8H18(l)+25O2(g) 16CO2(g)+18H2O(g)

Number of moles of gasoline (C8H18) can be determined form the density and molar mass C8H18 as shown below,

  n = 0.050 mL C8H18×0.760 g C8H181 mL C8H18×1 mol C8H18114.2 g C8H18=3.3×104mol C8H18

From the balanced equation, it is clear that 2 mol C8H18 reacts with 25 mol O2. Hence, number of moles of O2 reacted is determined as follows,

  nO2 = 3.3×104mol C8H18×25 mol O22 mol C8H18=0.0041 mol O2

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Mass of NO formed from N2 has to be calculated.

Concept Introduction:

Refer to (c)

Answer to Problem ISP

Mass of NO formed is 1.56 g

Explanation of Solution

Given information is shown below,

  P = 4012 mm Hg= 4012 mm Hg×1 atm760 mm Hg = 5.279e atmVcylinder = 485 mL = 0.485 LT = 1200 K

Number of moles of N2 is calculated using Ideal gas law as follows,

  PV=nN2RT(5.279 atm)(0.485 L)= nN2(0.0821 L.atm.mol1.K1)(1200 K)nN2=(5.279 atm)(0.485 L)(0.0821 L.atm.mol1.K1)(1200 K)=2.599×102mol N2

The reaction that shows the formation of NO from N2 is given below,

  N2(g)+O2(g) 2NO(g)

From the balanced equation, it is clear that 1 mol N2 reacts to give 2 mol NO. Hence, number of moles of NO formed is determined,

  n = 2.599×102mol N2×2 mol NO1 mol N2=0.052 mol NO

Mass of NO formed is determined as follows,

  Mass= No. of moles×Molar mass= 0.052 mol NO×30.01 g NO1 mol NO= 1.56 g NO

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Mass of N2O required has to be calculated.

Answer to Problem ISP

Mass of N2O required is 1.14 g.

Explanation of Solution

The reaction that shows the formation of NO from N2O is given below,

  N2O(g)+ 12O2(g) 2NO(g)

Number of moles of NO formed is 0.052 mol.

From the balanced equation, it is clear that 1 mol N2O reacts to give 2 mol NO. Hence, mass of N2O required is determined as follows,

  Mass = 0.052 mol NO×(1 mol N2O2 mol NO)×(44.01 g N2O1 mol N2O)= 1.14 g N2O

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Chapter 8 Solutions

Chemistry: The Molecular Science

Ch. 8.4 - Prob. 8.5PSPCh. 8.4 - Prob. 8.8CECh. 8.4 - Prob. 8.9CECh. 8.4 - Prob. 8.6PSPCh. 8.4 - Prob. 8.10CECh. 8.5 - Prob. 8.7PSPCh. 8.5 - Prob. 8.8PSPCh. 8.5 - Prob. 8.11ECh. 8.6 - Prob. 8.9PSPCh. 8.6 - Prob. 8.12CECh. 8.6 - Prob. 8.13ECh. 8.6 - Prob. 8.10PSPCh. 8.6 - Prob. 8.11PSPCh. 8.7 - Prob. 8.12PSPCh. 8.7 - Prob. 8.14ECh. 8.7 - Prob. 8.16CECh. 8.7 - Prob. 8.17ECh. 8.8 - Prob. 8.13PSPCh. 8.8 - Prob. 8.18ECh. 8.8 - Look up the van der Waals constants, b, for H2,...Ch. 8.11 - List as many natural sources of CO2 as you can,...Ch. 8.11 - Prob. 8.21ECh. 8.11 - Prob. 8.22CECh. 8.11 - Prob. 8.23CECh. 8.11 - Prob. 8.24CECh. 8.12 - Make these conversions for atmospheric...Ch. 8.12 - Prob. 8.25ECh. 8 - In a typical automobile engine, a gasoline...Ch. 8 - Prob. 1QRTCh. 8 - Prob. 2QRTCh. 8 - Prob. 3QRTCh. 8 - Prob. 4QRTCh. 8 - Prob. 5QRTCh. 8 - Prob. 6QRTCh. 8 - Prob. 7QRTCh. 8 - Prob. 8QRTCh. 8 - Prob. 9QRTCh. 8 - Prob. 10QRTCh. 8 - Prob. 11QRTCh. 8 - Prob. 12QRTCh. 8 - Prob. 13QRTCh. 8 - Prob. 14QRTCh. 8 - Prob. 15QRTCh. 8 - Prob. 16QRTCh. 8 - Prob. 17QRTCh. 8 - Prob. 18QRTCh. 8 - Some butane, the fuel used in backyard grills, is...Ch. 8 - Prob. 20QRTCh. 8 - Suppose you have a sample of CO2 in a gas-tight...Ch. 8 - Prob. 22QRTCh. 8 - Prob. 23QRTCh. 8 - Prob. 24QRTCh. 8 - A sample of gas occupies 754 mL at 22 C and a...Ch. 8 - Prob. 26QRTCh. 8 - Prob. 27QRTCh. 8 - Prob. 28QRTCh. 8 - Prob. 29QRTCh. 8 - Prob. 30QRTCh. 8 - Prob. 31QRTCh. 8 - Prob. 32QRTCh. 8 - Calculate the molar mass of a gas that has a...Ch. 8 - Prob. 34QRTCh. 8 - Prob. 35QRTCh. 8 - Prob. 36QRTCh. 8 - Prob. 37QRTCh. 8 - Prob. 38QRTCh. 8 - Prob. 39QRTCh. 8 - Prob. 40QRTCh. 8 - Prob. 41QRTCh. 8 - Prob. 42QRTCh. 8 - Prob. 43QRTCh. 8 - Prob. 44QRTCh. 8 - Prob. 45QRTCh. 8 - Prob. 46QRTCh. 8 - Prob. 47QRTCh. 8 - Prob. 48QRTCh. 8 - The build-up of excess carbon dioxide in the air...Ch. 8 - Prob. 50QRTCh. 8 - Prob. 51QRTCh. 8 - Prob. 52QRTCh. 8 - Prob. 53QRTCh. 8 - Prob. 54QRTCh. 8 - Prob. 55QRTCh. 8 - Benzene has acute health effects. For example, it...Ch. 8 - The mean fraction by mass of water vapor and cloud...Ch. 8 - Acetylene can be made by reacting calcium carbide...Ch. 8 - Prob. 59QRTCh. 8 - You are given two flasks of equal volume. Flask A...Ch. 8 - Prob. 61QRTCh. 8 - Prob. 62QRTCh. 8 - Prob. 63QRTCh. 8 - Prob. 64QRTCh. 8 - Prob. 65QRTCh. 8 - Prob. 66QRTCh. 8 - Prob. 67QRTCh. 8 - Prob. 68QRTCh. 8 - Prob. 69QRTCh. 8 - Prob. 70QRTCh. 8 - Prob. 71QRTCh. 8 - Prob. 72QRTCh. 8 - Prob. 73QRTCh. 8 - Prob. 74QRTCh. 8 - Prob. 75QRTCh. 8 - Prob. 76QRTCh. 8 - Prob. 77QRTCh. 8 - Prob. 78QRTCh. 8 - Prob. 79QRTCh. 8 - Prob. 80QRTCh. 8 - Prob. 81QRTCh. 8 - Prob. 82QRTCh. 8 - Prob. 83QRTCh. 8 - Prob. 84QRTCh. 8 - Prob. 85QRTCh. 8 - Name a favorable effect of the global increase of...Ch. 8 - Prob. 87QRTCh. 8 - Assume that limestone, CaCO3, is used to remove...Ch. 8 - Prob. 89QRTCh. 8 - Prob. 90QRTCh. 8 - Prob. 91QRTCh. 8 - Prob. 92QRTCh. 8 - Prob. 93QRTCh. 8 - Prob. 94QRTCh. 8 - Prob. 95QRTCh. 8 - Prob. 96QRTCh. 8 - Prob. 97QRTCh. 8 - Prob. 98QRTCh. 8 - Prob. 99QRTCh. 8 - Prob. 100QRTCh. 8 - Prob. 101QRTCh. 8 - Prob. 102QRTCh. 8 - Prob. 103QRTCh. 8 - Prob. 104QRTCh. 8 - Prob. 105QRTCh. 8 - Prob. 106QRTCh. 8 - Prob. 107QRTCh. 8 - Prob. 108QRTCh. 8 - Prob. 109QRTCh. 8 - Consider these four gas samples, all at the same...Ch. 8 - Prob. 111QRTCh. 8 - Prob. 112QRTCh. 8 - Prob. 113QRTCh. 8 - Prob. 114QRTCh. 8 - Prob. 115QRTCh. 8 - Prob. 116QRTCh. 8 - Prob. 117QRTCh. 8 - Prob. 118QRTCh. 8 - Prob. 119QRTCh. 8 - Prob. 120QRTCh. 8 - Prob. 121QRTCh. 8 - Prob. 122QRTCh. 8 - Prob. 123QRTCh. 8 - Prob. 124QRTCh. 8 - Prob. 125QRTCh. 8 - Prob. 126QRTCh. 8 - Prob. 127QRTCh. 8 - Prob. 128QRTCh. 8 - Prob. 129QRTCh. 8 - Prob. 8.ACPCh. 8 - Prob. 8.BCP
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