Calculate the molar flow rate of each element in the coal. Concept introduction: It is given that wet coal being used in a power plant at a rate of 8000 lb m / min has the following composition. Component Weight % (dry basis) Ash 7.2 Sulfur 3.5 Hydrogen 5.0 Carbon 75.2 Nitrogen 1.6 Oxygen 7.5 In addition, the wet coal contains 4.58 lb m H 2 O per lb m of dry coal. Units can be converted as, 1 l b m = 0.4536 k g Molecular weight of each element is as follows, Sulfur = 32 g/mol Hydrogen = 1 g/mol Carbon = 12 g/mol Nitrogen = 14 g/mol Oxygen = 16 g/mol

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Answer 1

Question

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

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Answer 2

Question

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

Expert Solution & Answer

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Answer 3

Question

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

Expert Solution & Answer

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Answer 4

Question

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

Expert Solution & Answer

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Answer 5

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

Answer 6

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

Answer 7

Chapter 3, Problem 3.30P

Interpretation Introduction

Interpretation:

Calculate the molar flow rate of each element in the coal.

Concept introduction:

It is given that wet coal being used in a power plant at a rate of 8000 lb_m / min has the following composition.

Component Weight % (dry basis)

Ash 7.2

Sulfur 3.5

Hydrogen 5.0

Carbon 75.2

Nitrogen 1.6

Oxygen 7.5

In addition, the wet coal contains 4.58 lb_m H₂ O per lb_m of dry coal.

Units can be converted as,

1 lbm = 0.4536 kg

Molecular weight of each element is as follows,

Sulfur = 32 g/mol

Hydrogen = 1 g/mol

Carbon = 12 g/mol

Nitrogen = 14 g/mol

Oxygen = 16 g/mol

Answer 8

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Answer 9

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Answer 11

Ch. 3 - Perform the following estimations without using a...Ch. 3 - Prob. 3.2P Ch. 3 - Prob. 3.3P Ch. 3 - Prob. 3.4P Ch. 3 - Prob. 3.5P Ch. 3 - Prob. 3.6P Ch. 3 - Prob. 3.7P Ch. 3 - Prob. 3.8P Ch. 3 - Prob. 3.9P Ch. 3 - A rectangular block of solid carbon (graphite)...

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Chapter 3 Solutions

Component	Weight % (dry basis)
Ash	7.2
Sulfur	3.5
Hydrogen	5.0
Carbon	75.2
Nitrogen	1.6
Oxygen	7.5