Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card
Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card
11th Edition
ISBN: 9781337128391
Author: Darrell Ebbing, Steven D. Gammon
Publisher: Cengage Learning
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Chapter 5, Problem 5.161QP

(a)

Interpretation Introduction

Interpretation:

Adding the N2 supplied in air to the power plant running on a fossil fuel having a formula C18H10S . The balanced combustion equation for a complex fuel having 120% stoichiometric combustion needs to be written.

Concept Introduction:

According to law of conservation of mass

Reactantsproducts

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

No of moles = mass of the substance×1.0 molmolar mass of the substance

(a)

Expert Solution
Check Mark

Answer to Problem 5.161QP

The power plant running on a fossil fuel having a formula C18H10S with added N2 gas the reactants and products of the reactions are balanced.

Explanation of Solution

Given,

Power plant driven by fossil fuel combustion generate substantial greenhouse gases (CO2andH2O) and gases that subsidizes to low air quality (e.gSO2) to evaluate the emission on a regular basis inert nitrogen gas is supplied with air to balance reactions.

The power plant running on a fossil fuel having the formula C18H10S

100%Combustion of C18H10S taken from problem 3.141 as,

2C18H10S+43O2+161.7N236CO2(g)+10H2O(l)+2SO2(g)+161.7N2(g)

Moles of Oxygen can be calculated as

No ofO2 moles = mass of the substance×1.0 molmolar mass of the substance

43molO2×3.76molN21.00molO2=161.7molN2

120%Combustion of C18H10S can calculated as

1.2×43=51.6molofO21.2×161.7=194.04molofN2

Balanced reaction for120%combustionis2C18H10S+51.6O2+194.04N236CO2(g)+10H2O(l)+2SO2(g)+194.04N2(g)+8.6O2(g)

Conclusion

The equation of power plant running on a fossil fuel gas is balanced.

(b)

Interpretation Introduction

Interpretation:

When deriving the balanced combustion reaction to effectively convert the N2O2 volume to a molar ratio, which law is used to define need to be identified.

Concept Introduction:

Boyle’s law:  According to Boyle’s law, pressure and volume are inversely proportional to each other at a given temperature.

PV=Constant                                          ...(1)or  Vα1P

Where,

P = Pressure

V = Volume

Charles law: According to Charles law, temperature and volume are directly proportional to each other at a constant pressure.

   VαT(ataconstantpressure)                      ...(2)orVT=C

Where,

T = Temperature

V = Volume.

Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

Vα n (ataconstantpressure)                    ...(3)orVn= k

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Ideal gas equation:

At a constant temperature (K) and pressure (P), the volume (v) occupied by the no of moles of any gas is known as ideal gas equation.

Ideal gas equation:

PV=nRT

(b)

Expert Solution
Check Mark

Answer to Problem 5.161QP

Avogadro’s law is used to explain the volume to mole ratio for the above given fossil fuel power plant operation.

Explanation of Solution

Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

Vα n (ataconstantpressure)orVn= k

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

Conclusion

The law used in the operation of fossil fuel power plant is identified.

(c)

Interpretation Introduction

Interpretation:

In fossil fuel power plant, assuming that the product water condenses from (b) calculate the volume/volume percentages for gases CO2,N2 and SO2 .

Concept Introduction:

1. According to law of conservation of mass

Reactantsproducts

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

No of moles = mass of the substance×1.0 molmolar mass of the substance

2. Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

Vα n (ataconstantpressure)orVn= k

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

(c)

Expert Solution
Check Mark

Answer to Problem 5.161QP

volumevolume%ofCO2=15%volumevolume%ofN2=80.6%volumevolume%ofSO2=0.83%volumevolume%ofO2=3.6%

Explanation of Solution

Chemical balanced equation

2C18H10S+51.6O2+194.04N236CO2(g)+10H2O(l)+2SO2(g)+194.04N2(g)+8.6O2(g)

According to Avogadro’s law

The product of gases are collected at the same temperature and pressure and volume are directly related in volumes.

The total gas volume = 36+2+194.0+8.6=240.6volumes

volumevolume%ofCO2=(36240.6)×100=15%volumevolume%ofN2=(194.0240.6)×100=80.6%volumevolume%ofSO2=(2240.6)×100=0.83%volumevolume%ofO2=(8.6240.6)×100=3.6%

Conclusion

The volume/volume ratio percentages of the gases CO2,N2 and SO2 are calculated as 15%,80.6%and0.83% respectively.

(d)

Interpretation Introduction

Interpretation:

In fossil fuel power plant, assuming that the product water vapor on wet basis from (c) calculate the volume/volume percentages for gases CO2,N2 and SO2 .

Concept Introduction:

1. According to law of conservation of mass

Reactantsproducts

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

No of moles = mass of the substance×1.0 molmolar mass of the substance

2. Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

Vα n (ataconstantpressure)orVn= k

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

(d)

Expert Solution
Check Mark

Answer to Problem 5.161QP

volumevolume%ofCO2=14.4%volumevolume%ofH2O=4.0%volumevolume%ofSO2=0.8%volumevolume%ofN2=77.4%volumevolume%ofO2=3.4%

Explanation of Solution

Chemical balanced equation

2C18H10S+51.6O2+194.04N236CO2(g)+10H2O(l)+2SO2(g)+194.04N2(g)+8.6O2(g)

According to Avogadro’s law

The product of gases are collected at the same temperature and pressure and volume are directly related in volumes.

The total gas volume = 36+2+10+8.6+194.0=250.6volumes

volumevolume%ofCO2=14.4%volumevolume%ofH2O=4.0%volumevolume%ofSO2=0.8%volumevolume%ofN2=77.4%volumevolume%ofO2=3.4%

Conclusion

The volume/volume ratio percentages of the gases CO2,H2O,SO2,N2 and O2 are calculated as 14.4%,4.00%,0.80%,77.4%,and3.4% respectively.

(e)

Interpretation Introduction

Interpretation:

Among the two fuels C18H10SandC32H28S keeping the air quality in mind which one is more likely to be used has to be identified.

(e)

Expert Solution
Check Mark

Answer to Problem 5.161QP

The fuel C32H28S can be used in the port.

Explanation of Solution

The fuel C32H28S can be used in the port because of the lower content of sulfur in C32H28S , it will give lesser concentration of SO2 in the Stack gas emissions.

Conclusion

The fuel lower content of sulfur is identified as C32H28S .

(f)

Interpretation Introduction

Interpretation:

In fossil fuel power plant, assuming that the product water vapor on wet basis from (d) calculate the volume/volume percentages for gases SO2 produced from the fuels C18H10SandC32H28S assuming that 120% stoichiometric combustion.

Concept Introduction:

1. According to law of conservation of mass

Reactantsproducts

The masses of the products and the reactants before and after the reaction are equal.

When a chemical reaction occurs number of reactant atoms are equal to number of product atoms are known as balanced chemical reaction.

No of moles = mass of the substance×1.0 molmolar mass of the substance

2. Avogadro’s law: According to Avogadro’s law, equal volumes of gases contain same number of molecules at a given temperature and volume.

Vα n (ataconstantpressure)orVn= k

Where,

n = no. of molecules (in moles)

V = Volume

k = constant

(f)

Expert Solution
Check Mark

Answer to Problem 5.161QP

volumevolume%ofSO2(C18H10S)=0.83%volumevolume%ofSO2(C32H12S)=0.55%

Explanation of Solution

Chemical balanced equation

2C18H10S+51.6O2+194.04N236CO2(g)+10H2O(l)+2SO2(g)+194.04N2(g)+8.6O2(g)

Similarly for C32H28S

2C32H18S+90O2+282N264CO2(g)+18H2O(l)+2SO2(g)+282N2(g)+15O2(g)

According to Avogadro’s law

The product of gases are collected at the same temperature and pressure and volume are directly related in volumes.

volumevolume%ofSO2(C18H10S)=(1363)×100=0.83%volumevolume%ofSO2(C32H12S)=(2363)×1000=0.55%

From volume/volume we can say that SO2 content in the air is reduced when we use the fuel with the formula C32H28S .

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Chapter 5 Solutions

Bundle: General Chemistry, Loose-leaf Version, 11th + OWLv2, 4 terms (24 months) Printed Access Card

Ch. 5.3 - Prob. 5.3CCCh. 5.4 - How many liters of chlorine gas, Cl2, can be...Ch. 5.5 - A 10.0-L flask contains 1.031 g O2 and 0.572 g CO2...Ch. 5.5 - A flask equipped with a valve contains 3.0 mol of...Ch. 5.5 - Prob. 5.11ECh. 5.6 - Prob. 5.5CCCh. 5.7 - What is the rms speed (in m/s) of a carbon...Ch. 5.7 - At what temperature do hydrogen molecules, H2,...Ch. 5.7 - Prob. 5.14ECh. 5.7 - If it takes 4.67 times as long for a particular...Ch. 5.7 - Prob. 5.6CCCh. 5.8 - Prob. 5.16ECh. 5.8 - Prob. 5.7CCCh. 5 - Prob. 5.1QPCh. 5 - Prob. 5.2QPCh. 5 - Prob. 5.3QPCh. 5 - Prob. 5.4QPCh. 5 - The volume occupied by a gas depends linearly on...Ch. 5 - Prob. 5.6QPCh. 5 - Prob. 5.7QPCh. 5 - Prob. 5.8QPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Under what conditions does the behavior of a real...Ch. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - A 1-liter container is filled with 2.0 mol Ar, 2.0...Ch. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - A 3.00-L flask containing 2.0 mol of O2 and 1.0...Ch. 5 - Prob. 5.33QPCh. 5 - Two identical He-filled balloons, each with a...Ch. 5 - You have a balloon that contains O2. What could...Ch. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - The barometric pressure measured outside an...Ch. 5 - Prob. 5.39QPCh. 5 - You fill a balloon with helium gas to a volume of...Ch. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - A McLeod gauge measures low gas pressures by...Ch. 5 - If 456 dm3 of krypton at 101 kPa and 21C is...Ch. 5 - A sample of nitrogen gas at 17C and 760 mmHg has a...Ch. 5 - Prob. 5.46QPCh. 5 - Helium gas, He, at 22C and 1.00 atm occupied a...Ch. 5 - Prob. 5.48QPCh. 5 - A vessel containing 39.5 cm3 of helium gas at 25C...Ch. 5 - A sample of 62.3 cm3 of argon gas at 18C was...Ch. 5 - A bacterial culture isolated from sewage produced...Ch. 5 - Pantothenic acid is a B vitamin. Using the Dumas...Ch. 5 - In the presence of a platinum catalyst, ammonia,...Ch. 5 - Methanol, CH3OH, can be produced in industrial...Ch. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - A cylinder of oxygen gas contains 91.3 g O2. If...Ch. 5 - In an experiment, you fill a heavy-walled 6.00-L...Ch. 5 - Prob. 5.59QPCh. 5 - According to your calculations, a reaction should...Ch. 5 - Prob. 5.61QPCh. 5 - A 2.50-L flask was used to collect a 5.65-g sample...Ch. 5 - What is the density of ammonia gas, NH3, at 31C...Ch. 5 - Calculate the density of hydrogen sulfide gas,...Ch. 5 - Butane, C4H10, is an easily liquefied gaseous...Ch. 5 - Chloroform, CHCl3, is a volatile (easily...Ch. 5 - A chemist vaporized a liquid compound and...Ch. 5 - You vaporize a liquid substance at 100C and 755...Ch. 5 - A 2.56-g sample of a colorless liquid was...Ch. 5 - A 2.30-g sample of white solid was vaporized in a...Ch. 5 - Ammonium chloride, NH4Cl, is a while solid. 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