The chemical equations for the combustion of propane and ethyne have to be written. Concept Introduction: A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons. Complete combustion occurs if enough oxygen is present. Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

Question

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Answer 1

Question

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.

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Answer 2

Question

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.

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Answer 3

Question

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.

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Answer 4

Question

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.

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Answer 5

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.

Answer 6

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Answer 7

Chapter 11, Problem 11.47E

(a)

Interpretation Introduction

Interpretation:

The chemical equations for the combustion of propane and ethyne have to be written.

Concept Introduction:

A hydrocarbon undergoes complete combustion to produce carbon dioxide and water. This is chemical property of all hydrocarbons.

Complete combustion occurs if enough oxygen is present.

Incomplete combustion occurs, if there is not enough oxygen present. Incomplete combustion results in the formation of toxic gases.

Answer 8

(a)

Expert Solution

Answer to Problem 11.47E

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The chemical equation for complete combustion reaction of ethyne (Acetylene) is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The incomplete combustion reaction of propane results in the formation of carbon, carbon monoxide, carbon dioxide and water. The chemical equation for incomplete combustion reaction of propane is,

$2C3H8(g)+7O2(g)→2C(s)+2CO(g)+2CO2(g)+8H2O(g)$

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Answer 13

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The enthalpy of combustion for each fuel per gram and per mole has to be calculated.

Concept Introduction:

Hess Law:

The total enthalpy change during the complete course of a chemical reaction is same whether the reaction is made in one step or in several steps. The enthalpy of reaction is calculated as,

$ΔHoreaction=ΔHof(products)-ΔHof(reactants)$

Answer 15

(b)

Expert Solution

Explanation of Solution

Combustion reaction of Propane:

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

Enthalpy of combustion of propane:

Heat of formation of propane= $-104 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of propane is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((3)(-394kJ/mole)+(4)(-241.8kJ/mole))-((1)(-104kJ/mole)+(5)(0))ΔHocombustion=-1182kJ/mole+(-967.2kJ/mole)-(-104kJ/mole)ΔHocombustion=-2045.2kJ/mole$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

The enthalpy of combustion of propane is $-2045.2 kJ$ per 44 grams of propane.

Combustion reaction of Acetylene:

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

Enthalpy of combustion of acetylene:

Heat of formation of acetylene= $226.7 kJ/mole$

Heat of formation of carbon dioxide gas= $-394 kJ/mole$

Heat of formation of water= $-241.8 kJ/mole$

The enthalpy of combustion of acetylene is,

$ΔHocombustion=ΔHof(products)-ΔHof(reactants)ΔHocombustion=((4)(-394kJ/mole)+(2)(-241.8kJ/mole))-((2)(226.7 kJ/mole)+(5)(0))ΔHocombustion=-1576 kJ/mole+(-483.6 kJ/mole)-(453.4 kJ/mole)ΔHocombustion=-2513 kJ/mole$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of acetylene is $-2513 kJ$ per 26.04 grams of acetylene.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.

Answer 20

(c)

Interpretation Introduction

Interpretation:

The reason why propane torches are not used for wielding has to be calculated.

Answer 21

(c)

Expert Solution

Explanation of Solution

Propane undergoes complete combustion reaction to form carbon dioxide and water.

The chemical equation for complete combustion reaction of propane is,

$C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)$

The enthalpy of combustion of propane is $-2045.2 kJ/mole.$

Acetylene (or) ethyne undergoes complete combustion reaction to form carbon dioxide and water vapor as their products. The chemical equation for complete combustion reaction of acetylene is,

$2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)$

The enthalpy of combustion of acetylene is $-2513 kJ/mole.$

The enthalpy of combustion of propane is higher than the enthalpy of combustion of acetylene, and hence it cannot be used for wielding purposes. It also produces a lower flame temperature and the pre-heat time is more. Propane cannot be used for gas wielding because it does not have a reducing zone.