Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
Question
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Chapter 5, Problem 5.139P

(a)

Interpretation Introduction

Interpretation:

The mass of calcium sulfate from the scrubbing 4 GL of flue gas is to be calculated.

Concept introduction:

The expression to calculate the partial pressure of the gas is as follows:

Pgas=XgasPtotal

Here, Pgas is the partial pressure of the gas, Xgas is the mole fraction of the gas and Ptotal is the total pressure.

The expression to calculate the mole fraction of the gas is as follows,

Xgas=Moles of gasTotal moles

Here, Xgas is the mole fraction of the gas.

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(a)

Expert Solution
Check Mark

Answer to Problem 5.139P

The mass of calcium sulfate from the scrubbing 4 GL of flue gas is 4 kg.

Explanation of Solution

The concentration of SO2 is 1000 times higher than its mole fraction in clear dry air that is 2×1010. Thus, the mole fraction of SO2 equals to 2×107.

The expression to calculate the volume of SO2 is as follows:

Volume of SO2 removed=[(mole fraction of SO2)(volume of flue gas)(percentage of SO2 removes)]        (1)

Substitute the value 2×107 for mole fraction of SO2, 4 GL for the volume of flue gas and 95 % for a percentage of SO2 removes in the equation (1).

Volume of SO2 removed=[(2×107)(4 GL(109 L1 GL))(95 %100 %)]=760 L

The formula to convert °C to Kelvin is:

T(K)=T(°C)+273.15        (2)

Substitute 25 °C for T(°C) in equation (5).

T(K)=25 °C+273.15=298.15 K

The expression to calculate the moles of the SO2 is as follows,

PV=nRT        (3)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of SO2 and R is the gas constant.

Rearrange the equation (3) to calculate n as follows,

n=PVRT        (4)

Substitute the value 1.0 atm for P, 298.15 K for T, 760 L for V and 0.0821 LatmmolK for R in the equation (4).

n=(760 L)(1.0 atm)(0.0821 LatmmolK)(298.15 K)=31.04814 mol

The equation for the reaction of CaCO3 with SO2 is as follows:

CaCO3(s) + SO2(g)CaSO3(s)+CO2(g)        (5)

The equation for the reaction of CaSO3 with O2 is as follows:

2CaSO3(s) + O2(g)2CaSO4(s)        (6)

From the equation (5), one mole of the SO2 will be formed one mole of CaSO3. From the equation (6), two moles of the CaSO3 will be formed two moles of CaSO4. Thus, the moles of CaSO4 is calculated from SO2 as follows:

Moles of CaSO4=[(Moles of SO2)(1 mol CaSO41 mol SO2)]        (7)

Substitute the value 31.04814 mol for moles of SO2 in the equation (7).

Moles of CaSO4=[(31.04814 mol)(1 mol CaSO41 mol SO2)]=31.04814 mol

The expression to calculate the mass of CaSO4 is as follows:

Moles of CaSO4=Mass of CaSO4Molar mass CaSO4        (8)

Rearrange the expression to calculate the mass of CaSO4 in the equation (8)

Mass of CaSO4=(Moles of CaSO4)(Molar mass CaSO4)        (9)

Substitute the value 31.04814 mol for the mass of CaSO4 and 136.14 g/mol for the molar mass of CaSO4 in the equation (9).

Mass of CaSO4=(31.04814 mol)(136.14 g/mol)(1 kg103g)=4.2269 kg=4 kg

Conclusion

The mass of calcium sulfate from the scrubbing 4 GL of flue gas is 4 kg.

(b)

Interpretation Introduction

Interpretation:

The volume of air at 25 °C and 1.0 atm is needed to react with all calcium sulfite is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

PV=nRT

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the gas and R is the gas constant.

(b)

Expert Solution
Check Mark

Answer to Problem 5.139P

The volume of air at 25 °C and 1.0 atm is 2000 L.

Explanation of Solution

From the equation (6), one mole of the O2 will be formed two moles of CaSO4. Thus, the moles of O2 is calculated from CaSO4 as follows:

Moles of O2=[(Moles of CaSO4)(1 mol O22 mol CaSO4)]        (10)

Substitute the value 31.04814 mol for moles of CaSO4 in the equation (10).

Moles of O2=[(31.04814 mol)(1 mol O22 mol CaSO4)]=15.52407 mol

The formula to convert °C to Kelvin is:

T(K)=T(°C)+273.15        (11)

Substitute 25 °C for T(°C) in equation (11).

T(K)=25 °C+273.15=298.15 K

The expression to calculate the volume of O2 is as follows,

PV=nRT        (12)

Here, P is the pressure, V is the volume, T is the temperature, n is the mole of the O2 and R is the gas constant.

Rearrange the equation (7) to calculate V as follows,

V=nRTP        (13)

Substitute the value 1.0 atm for P, 298.15 K for T, 15.52407 mol for n and 0.0821 LatmmolK for R in the equation (13).

V=(15.52407 mol)(0.0821 LatmmolK)(298.15 K)1 atm=380 L

The expression to calculate the volume of air is as follows:

Volume of air=(Volume of O2)(Mole fraction of O2)        (14)

Substitute the value 0.209 for mole fraction of O2, 380 L for the volume of O2 in the equation (14).

Volume of air=(380 L)(0.209)=1818.18 L2000 L

Conclusion

The volume of air at 25 °C and 1.0 atm is 2000 L.

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Chapter 5 Solutions

Chemistry: The Molecular Nature of Matter and Change

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