Chapter 3, Problem 3.141QP

A power plant is driven by the combustion of a complex fossil fuel having the formula C₁₁H₇S. Assume the air supply is composed of only N₂ and O₂ with a molar ratio of 3.76:1.00, and the N₂ remains unreacted. In addition to the water produced, the fuel’s C is completely combusted to CO₂ and its sulfur content is converted to SO₂. In order to evaluate gases emitted at the exhaust stacks for environmental regulation purposes, the nitrogen supplied with the air must also be included in the balanced reactions.

1. a Including the N₂ supplied m the air, write a balanced combustion equation for the complex fuel assuming 100% stoichiometric combustion (i.e., when there is no excess oxygen in the products and the only C-containing product is CO₂). Except in the case of N₂, use only integer coefficients.
2. b Including N₂ supplied in the air, write a balanced combustion equation for the complex fuel assuming 120% stoichiometric combustion (i.e., when excess oxygen is present in the products and the only C-containing product is CO₂). Except in the case of use only integer coefficients
3. c Calculate the minimum mass (in kg) of air required to completely combust 1700 kg of C₁₁H₇S.
4. d Calculate the air/fuel mass ratio, assuming 100% stoichiometric combustion.
5. e Calculate the air/fuel mass ratio, assuming 120% stoichiometric combustion.

Interpretation Introduction

Interpretation:

The balanced equation for the combustion reaction should be written.

Concept introduction:

Balanced equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

The equation for a reaction, which has same number of atoms and charge of the ions in both reactants and product sides is known as balanced equation.

Answer to Problem 3.141QP

The balanced equation for the combustion reaction is,

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{55O}}_{\text{2}}\text{+55}\text{(3}\text{.76}{\text{)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}{\text{)N}}_{\text{2}}$

Explanation of Solution

The balance the given combustion reaction.

Given,

${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$

Supplied air with ratio is ${\text{O}}_{\text{2}}\text{and }{\text{N}}_{\text{2}}=3.\text{76}\text{:}\text{l}\text{.0}$

Products of combustion reaction is ${\text{CO}}_{\text{2}}{\text{, H}}_{\text{2}}{\text{O and SO}}_{\text{2}}$

Frist arrange the equation for the combustion reaction is,

${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O +SO}}_{\text{2}}$

Add the 11 coefficient for balancing carbon atom of ${\text{CO}}_{\text{2}}$

${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{11CO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O +SO}}_{\text{2}}$

Multiply the whole reaction 2 for balancing Hydrogen atom of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$

${\text{2C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{22CO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O +2SO}}_{\text{2}}$

Add the 7 coefficient for balancing carbon atom of ${\text{H}}_{\text{2}}\text{O}$

${\text{2C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{22CO}}_{\text{2}}{\text{+7H}}_{\text{2}}{\text{O +2SO}}_{\text{2}}$

In product side has 55 Oxygen atoms so add $\frac{55}{2}$ in reactant side.

${\text{2C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}\frac{55}{2}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{22CO}}_{\text{2}}{\text{+7H}}_{\text{2}}{\text{O +2SO}}_{\text{2}}$

Modify the $\frac{55}{2}$ into whole number by multiplying the reaction in to two

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}55{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}$

Add the nu reacted ${\text{N}}_{\text{2}}$ and its coefficient. The ratio of air is ${\text{O}}_{\text{2}}\text{and }{\text{N}}_{\text{2}}=3.\text{76}\text{:}\text{l}\text{.0}$ therefore the coefficient ${\text{N}}_{\text{2}}$ is $\text{55×}\text{3}\text{.76}$ and add this into both side of the reaction to get the balanced equation for given combustion reaction.

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{55O}}_{\text{2}}\text{+55}\text{(3}\text{.76}{\text{)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}{\text{)N}}_{\text{2}}$

Conclusion

The balanced equation for the combustion reaction was written.

Interpretation Introduction

Interpretation:

The balanced equation for the $\text{120}\text{\%}$ combustion reaction should be written.

Concept introduction:

Balanced equation:

A balanced chemical equation is an equation which contains same elements in same number on both the sides (reactant and product side) of the chemical equation thereby obeying the law of conservation of mass.

The equation for a reaction, which has same number of atoms and charge of the ions in both reactants and product sides is known as balanced equation.

Answer to Problem 3.141QP

The balanced equation for the $\text{120}\text{\%}$ combustion reaction is,

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}66{\text{O}}_{\text{2}}\text{+248}{\text{.16 N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+248}\text{.16}{\text{N}}_{\text{2}}+11{\text{O}}_{\text{2}}$

Explanation of Solution

The balance the given combustion reaction.

Given,

${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$

Supplied air with ratio is ${\text{O}}_{\text{2}}\text{and }{\text{N}}_{\text{2}}=3.\text{76}\text{:}\text{l}\text{.0}$

Products of combustion reaction is ${\text{CO}}_{\text{2}}{\text{, H}}_{\text{2}}{\text{O and SO}}_{\text{2}}$

Frist arrange the equation for the combustion reaction is,

${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O +SO}}_{\text{2}}$

Add the 11 coefficient for balancing carbon atom of ${\text{CO}}_{\text{2}}$

${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{11CO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O +SO}}_{\text{2}}$

Multiply the whole reaction 2 for balancing Hydrogen atom of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$

${\text{2C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{22CO}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O +2SO}}_{\text{2}}$

Add the 7 coefficient for balancing carbon atom of ${\text{H}}_{\text{2}}\text{O}$

${\text{2C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{22CO}}_{\text{2}}{\text{+7H}}_{\text{2}}{\text{O +2SO}}_{\text{2}}$

In product side has 55 Oxygen atoms so add $\frac{55}{2}$ in reactant side.

${\text{2C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}\frac{55}{2}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{22CO}}_{\text{2}}{\text{+7H}}_{\text{2}}{\text{O +2SO}}_{\text{2}}$

Modify the $\frac{55}{2}$ into whole number by multiplying the reaction in to two

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}55{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}$

Add the nu reacted ${\text{N}}_{\text{2}}$ and its coefficient. The ratio of air is ${\text{O}}_{\text{2}}\text{and }{\text{N}}_{\text{2}}=3.\text{76}\text{:}\text{l}\text{.0}$ therefore the coefficient ${\text{N}}_{\text{2}}$ is $\text{55×}\text{3}\text{.76}$ and add this into both side of the reaction to get the balanced equation for given combustion reaction.

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{55O}}_{\text{2}}\text{+55}\text{(3}\text{.76}{\text{)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}{\text{)N}}_{\text{2}}$

For the above $\text{120}\text{\%}$ combustion reaction, the excess of Oxygen is add in to Carbon containing products only.

Hence, the above balanced equation, ${\text{O}}_{\text{2}}\text{and }{\text{N}}_{\text{2}}=3.\text{76}\text{:}\text{l}\text{.0}$ is multiplied by 1.20,

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}\text{55}\text{(1}\text{.20)}{\text{O}}_{\text{2}}\text{+55}\text{(3}\text{.76}\text{)}\text{(1}{\text{.20)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}\text{)}\text{(1}{\text{.20)N}}_{\text{2}}+\text{(}{\text{.20)O}}_{\text{2}}$

Simplifying the above equation to give the balanced equation for the balanced equation for the $\text{120}\text{\%}$ combustion reaction.

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}66{\text{O}}_{\text{2}}\text{+248}{\text{.16 N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+248}\text{.16}{\text{N}}_{\text{2}}+11{\text{O}}_{\text{2}}$

Conclusion

The balanced equation for the $\text{120}\text{\%}$ combustion reaction was written.

Interpretation Introduction

Interpretation:

The minimum mass (in kg) of air required to completely combust $\text{1700 kg}$ of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ should be calculated.

Answer to Problem 3.141QP

The minimum mass (in kg) of air required to completely combust $\text{1700 kg}$ of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ is $\text{1}{\text{.9×10}}^{\text{4}}\text{kg}$.

Explanation of Solution

The balance the given combustion reaction.

The balanced equation for given $\text{100\%}$ combustion reaction is,

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{55O}}_{\text{2}}\text{+55}\text{(3}\text{.76}{\text{)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}{\text{)N}}_{\text{2}}$

The mass of are id divided by fuel mass to give an air/fuel mass ratio of $\text{100\%}$ combustion reaction.

Molar mass of ${\text{O}}_{\text{2}}$ is $\text{32}\text{.0}\text{g}$

Molar mass of ${\text{N}}_{\text{2}}$ is $\text{28}\text{.02}\text{g}$

Molar mass of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ is $\text{171}\text{.23 g}$

The air/fuel mass ratio of $\text{100\%}$ combustion reaction is,

$\begin{array}{l}\text{= }\frac{\text{Mass}\text{of}\text{air}}{\text{Mass}\text{of}\text{fuel}}\\ \text{=}\frac{\text{55}\text{mole}{\text{O}}_{\text{2}}\left(\frac{\text{32}\text{.0}\text{g}{\text{O}}_{\text{2}}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}\right)\text{+206}\text{.8}\text{mole}{\text{N}}_{\text{2}}\left(\frac{\text{28}\text{.02}\text{g}{\text{N}}_{\text{2}}}{\text{mole}{\text{N}}_{\text{2}}}\right)}{\text{4}\text{mole}\left(\frac{\text{171}\text{.23}\text{g}{\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}}{\text{1}\text{mol}{\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}}\right)}\\ \text{=}\frac{\text{7554}\text{g}\text{air}}{\text{684}\text{.9}\text{g}\text{fuel}}\\ \text{=}\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.\end{array}$

To calculate the minimum mass (in kg) of air required to completely combust $\text{1700 kg}$ of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$.

From the above calculation, the air/fuel mass ratio of $\text{100\%}$ combustion reaction is $\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.$

Therefore the air/fuel mass ratio of $\text{100\%}$ combustion is multiplied with given mass $\text{(1700 kg)}$ to give the minimum mass (in kg) of air required to completely combust $\text{1700 kg}$ of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$.

$\begin{array}{l}\text{=1700}\text{kg}\text{fuel}\text{×}\frac{\text{11}\text{.03}\text{g}\text{of}\text{air}}{\text{1}\text{g}\text{of}\text{fuel}}\\ \text{=}\text{1}{\text{.9×10}}^{\text{4}}\text{kg}\end{array}$

The minimum mass (in kg) of air required to completely combust $\text{1700 kg}$ of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ is $\text{1}{\text{.9×10}}^{\text{4}}\text{kg}$.

Conclusion

The minimum mass (in kg) of air required to completely combust $\text{1700 kg}$ of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ was calculated.

Interpretation Introduction

Interpretation:

The air/fuel mass ratio of $\text{100\%}$ combustion reaction should be calculated.

Answer to Problem 3.141QP

The air/fuel mass ratio of $\text{100\%}$ combustion reaction is $\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.$

Explanation of Solution

The balance the given combustion reaction.

The balanced equation for given $\text{100\%}$ combustion reaction is,

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{55O}}_{\text{2}}\text{+55}\text{(3}\text{.76}{\text{)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}{\text{)N}}_{\text{2}}$

The mass of are id divided by fuel mass to give an air/fuel mass ratio of $\text{100\%}$ combustion reaction.

Molar mass of ${\text{O}}_{\text{2}}$ is $\text{32}\text{.0}\text{g}$

Molar mass of ${\text{N}}_{\text{2}}$ is $\text{28}\text{.02}\text{g}$

Molar mass of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ is $\text{171}\text{.23 g}$

The air/fuel mass ratio of $\text{100\%}$ combustion reaction is,

$\begin{array}{l}\text{= }\frac{\text{Mass}\text{of}\text{air}}{\text{Mass}\text{of}\text{fuel}}\\ \text{=}\frac{\text{55}\text{mole}{\text{O}}_{\text{2}}\left(\frac{\text{32}\text{.0}\text{g}{\text{O}}_{\text{2}}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}\right)\text{+206}\text{.8}\text{mole}{\text{N}}_{\text{2}}\left(\frac{\text{28}\text{.02}\text{g}{\text{N}}_{\text{2}}}{\text{mole}{\text{N}}_{\text{2}}}\right)}{\text{4}\text{mole}\left(\frac{\text{171}\text{.23}\text{g}{\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}}{\text{1}\text{mol}{\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}}\right)}\\ \text{=}\frac{\text{7554}\text{g}\text{air}}{\text{684}\text{.9}\text{g}\text{fuel}}\\ \text{=}\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.\end{array}$

The masses of air and fuel are plugged above equation to give the air/fuel mass ratio of $\text{100\%}$ combustion reaction.

The air/fuel mass ratio of $\text{100\%}$ combustion reaction is $\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.$

Conclusion

The air/fuel mass ratio of $\text{100\%}$ combustion reaction was calculated.

Interpretation Introduction

Interpretation:

The air/fuel mass ratio of $\text{100\%}$ combustion reaction should be calculated.

Answer to Problem 3.141QP

The air/fuel mass ratio of $\text{120\%}$ combustion reaction is $\text{13}\text{.0}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.$

Explanation of Solution

The balance the given combustion reaction.

The balanced equation for given $\text{100\%}$ combustion reaction is,

${\text{4C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S+}{\text{55O}}_{\text{2}}\text{+55}\text{(3}\text{.76}{\text{)N}}_{\text{2}}\stackrel{}{\to }{\text{44CO}}_{\text{2}}{\text{+14H}}_{\text{2}}{\text{O +4SO}}_{\text{2}}\text{+55(}\text{3}\text{.76}{\text{)N}}_{\text{2}}$

The mass of are id divided by fuel mass to give an air/fuel mass ratio of $\text{100\%}$ combustion reaction.

Molar mass of ${\text{O}}_{\text{2}}$ is $\text{32}\text{.0}\text{g}$

Molar mass of ${\text{N}}_{\text{2}}$ is $\text{28}\text{.02}\text{g}$

Molar mass of ${\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}$ is $\text{171}\text{.23 g}$

The air/fuel mass ratio of $\text{100\%}$ combustion reaction is,

$\begin{array}{l}\text{= }\frac{\text{Mass}\text{of}\text{air}}{\text{Mass}\text{of}\text{fuel}}\\ \text{=}\frac{\text{55}\text{mole}{\text{O}}_{\text{2}}\left(\frac{\text{32}\text{.0}\text{g}{\text{O}}_{\text{2}}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}\right)\text{+206}\text{.8}\text{mole}{\text{N}}_{\text{2}}\left(\frac{\text{28}\text{.02}\text{g}{\text{N}}_{\text{2}}}{\text{mole}{\text{N}}_{\text{2}}}\right)}{\text{4}\text{mole}\left(\frac{\text{171}\text{.23}\text{g}{\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}}{\text{1}\text{mol}{\text{C}}_{\text{11}}{\text{H}}_{\text{7}}\text{S}}\right)}\\ \text{=}\frac{\text{7554}\text{g}\text{air}}{\text{684}\text{.9}\text{g}\text{fuel}}\\ \text{=}\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.\end{array}$

The masses of air and fuel are plugged above equation to give the air/fuel mass ratio of $\text{100\%}$ combustion reaction.

The air/fuel mass ratio of $\text{100\%}$ combustion reaction is $\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.$

To calculate the air/fuel mass ratio of $\text{120\%}$ combustion reaction.

The calculated air/fuel mass ratio of $\text{100\%}$ combustion reaction is multiplied with 1.2 to give the air/fuel mass ratio of $\text{120\%}$ combustion reaction,

$\begin{array}{l}\text{For }\text{100}\text{\%}\text{=}\frac{\text{7554}\text{g}\text{air}}{\text{684}\text{.9}\text{g}\text{fuel}}\\ \text{=}\text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.\\ \text{For }\text{120}\text{\%}\text{=}1.2\times \text{11}\text{.03}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.\\ =\text{13}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.\end{array}$

The air/fuel mass ratio of $\text{120\%}$ combustion reaction is $\text{13}\text{g}\raisebox{1ex}{$\text{air}$}\!\left/ \!\raisebox{-1ex}{$\text{fuel}$}\right.$

Conclusion

The air/fuel mass ratio of $\text{120\%}$ combustion reaction was calculated.