Chapter 15.7, Problem 38P

To determine

The mass fraction and the apparent molecular weight of the products and the mass of air required per unit mass of fuel burned.

Answer to Problem 38P

The mass fraction of carbon dioxide is $0.2441$, carbon monoxide is $0.0082$, water is $0.0595$, sulphur dioxide is $0.0032$, nitrogen is $0.6849$, the apparent molecular weight of the products is $\text{29}\text{.71}\text{kg}/\text{kmol}$ and the mass of air required per unit mass of fuel burned is $\text{8}\text{.27}\text{kg}\text{air}/\text{kg}\text{fuel}$.

Explanation of Solution

Express the total mass of the coal when the ash is substituted.

$m_{\text{total}}=100-m_{\text{ash}}$ (I)

Here, mass of ash is $m_{\text{ash}}$.

Express the mass fraction of carbon.

$m{f}_{\text{C}}=\frac{m_{\text{C}}}{m_{\text{total}}}$ (II)

Here, mass of carbon is $m_{\text{C}}$.

Express the mass fraction of hydrogen.

$m{f}_{{\text{H}}_2}=\frac{m_{{\text{H}}_2}}{m_{\text{total}}}$ (III)

Here, mass of hydrogen is $m_{{\text{H}}_2}$.

Express the mass fraction of oxygen.

$m{f}_{{\text{O}}_2}=\frac{m_{{\text{O}}_2}}{m_{\text{total}}}$ (IV)

Here, mass of oxygen is $m_{{\text{O}}_2}$.

Express the mass fraction of nitrogen.

$m{f}_{{\text{N}}_2}=\frac{m_{{\text{N}}_2}}{m_{\text{total}}}$ (V)

Here, mass of nitrogen is $m_{{\text{N}}_2}$.

Express the mass fraction of sulphur.

$m{f}_{\text{S}}=\frac{m_{\text{S}}}{m_{\text{total}}}$ (VI)

Here, mass of sulphur is $m_{\text{S}}$.

Express the number of moles of carbon.

$N_{\text{C}}=\frac{m{f}_{\text{C}}}{M_{\text{C}}}$ (VII)

Here, molar mass of carbon is $M_{\text{C}}$.

Express the number of moles of hydrogen.

$N_{{\text{H}}_{\text{2}}}=\frac{m{f}_{{\text{H}}_{\text{2}}}}{M_{{\text{H}}_{\text{2}}}}$ (VIII)

Here, molar mass of hydrogen is $M_{{\text{H}}_{\text{2}}}$.

Express the number of moles of oxygen.

$N_{{\text{O}}_{\text{2}}}=\frac{m{f}_{{\text{O}}_{\text{2}}}}{M_{{\text{O}}_{\text{2}}}}$ (IX)

Here, molar mass of oxygen is $M_{{\text{O}}_{\text{2}}}$.

Express the number of moles of nitrogen.

$N_{{\text{N}}_{\text{2}}}=\frac{m{f}_{{\text{N}}_{\text{2}}}}{M_{{\text{N}}_{\text{2}}}}$ (X)

Here, molar mass of nitrogen is $M_{{\text{N}}_{\text{2}}}$.

Express the number of moles of sulphur.

$N_{\text{S}}=\frac{m{f}_{\text{S}}}{M_{\text{S}}}$ (XI)

Here, molar mass of sulphur is $M_{\text{S}}$.

Express the total number of moles.

$N_m=N_{\text{C}}+N_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}+N_{\text{S}}$ (XII)

Express the mole fraction of carbon.

$y_{\text{C}}=\frac{N_{\text{C}}}{N_m}$ (XIII)

Express the mole fraction of hydrogen.

$y_{{\text{H}}_{\text{2}}}=\frac{N_{{\text{H}}_{\text{2}}}}{N_m}$ (XIV)

Express the mole fraction of oxygen.

$y_{{\text{O}}_{\text{2}}}=\frac{N_{{\text{O}}_{\text{2}}}}{N_m}$ (XV)

Express the mole fraction of nitrogen.

$y_{{\text{N}}_{\text{2}}}=\frac{N_{{\text{N}}_{\text{2}}}}{N_m}$ (XVI)

Express the mole fraction of sulphur.

$y_{\text{S}}=\frac{N_{\text{S}}}{N_m}$ (XVII)

Express the total molar mass of the products.

$m_{\text{total}}=N_{{\text{CO}}_{\text{2}}}{M}_{{\text{CO}}_{\text{2}}}+N_{\text{CO}}{M}_{\text{CO}}+N_{{\text{H}}_{\text{2}}\text{O}}{M}_{{\text{H}}_{\text{2}}\text{O}}+N_{{\text{SO}}_{\text{2}}}{M}_{{\text{SO}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}{M}_{{\text{N}}_{\text{2}}}$ (XVIII)

Here, number of moles of carbon dioxide, carbon monoxide, water, sulphur dioxide, and nitrogen is $N_{{\text{CO}}_{\text{2}}},N_{\text{CO}},N_{{\text{H}}_{\text{2}}\text{O}},N_{{\text{SO}}_{\text{2}}}\text{and}{N}_{{\text{N}}_{\text{2}}}$ respectively, molar masses of carbon dioxide, carbon monoxide, water, sulphur dioxide, and nitrogen is $M_{{\text{CO}}_{\text{2}}},M_{\text{CO}},M_{{\text{H}}_{\text{2}}\text{O}},M_{{\text{SO}}_{\text{2}}}\text{and}{M}_{{\text{N}}_{\text{2}}}$ respectively.

Express the mole fraction of carbon dioxide.

$m{f}_{{\text{CO}}_2}=\frac{N_{{\text{CO}}_{\text{2}}}{M}_{{\text{CO}}_2}}{m_{\text{total}}}$ (XIX)

Here, molar mass of carbon dioxide is $M_{{\text{CO}}_{\text{2}}}$.

Express the mole fraction of carbon monoxide.

$m{f}_{\text{CO}}=\frac{N_{\text{CO}}{M}_{\text{CO}}}{m_{\text{total}}}$ (XX)

Here, molar mass of carbon monoxide is $M_{\text{CO}}$.

Express the mole fraction of water.

$m{f}_{{\text{H}}_2\text{O}}=\frac{N_{{\text{H}}_2\text{O}}{M}_{{\text{H}}_2\text{O}}}{m_{\text{total}}}$ (XXI)

Here, molar mass of water is $M_{{\text{H}}_2\text{O}}$.

Express the mole fraction of sulphur dioxide.

$m{f}_{{\text{SO}}_2}=\frac{N_{{\text{SO}}_2}{M}_{{\text{SO}}_2}}{m_{\text{total}}}$ (XXII)

Here, molar mass of sulphur dioxide is $M_{{\text{SO}}_2}$.

Express the mole fraction of nitrogen.

$m{f}_{{\text{N}}_2}=\frac{N_{{\text{N}}_2}{M}_{{\text{N}}_2}}{m_{\text{total}}}$ (XXIII)

Here, molar mass of nitrogen is $M_{{\text{N}}_2}$.

Express the total number of moles of product.

$N_{m,P}=N_{{\text{CO}}_{\text{2}}}+N_{\text{CO}}+N_{{\text{H}}_{\text{2}}\text{O}}+N_{{\text{SO}}_{\text{2}}}+N_{{\text{N}}_{\text{2}}}$ (XXIV)

Express the apparent molecular weight of the product gas.

$M_m=\frac{m_{\text{total}}}{N_{m,P}}$ (XXV)

Express the air-fuel mass ratio.

$\text{FA}=\frac{0.6403N_{\text{air}}{M}_{\text{air}}}{N_{\text{C}}{M}_{\text{C}}+N_{{\text{H}}_{\text{2}}}{M}_{{\text{H}}_{\text{2}}}+N_{{\text{O}}_2}{M}_{{\text{O}}_2}+N_{{\text{N}}_2}{M}_{{\text{N}}_2}+N_{\text{S}}{M}_{\text{S}}}$ (XXVI)

Conclusion:

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

$\begin{array}{l}{M}_{\text{C}}=12\text{kg}/\text{kmol}\\ M_{{\text{H}}_2}=2\text{kg}/\text{kmol}\\ M_{{\text{O}}_{\text{2}}}=32\text{kg}/\text{kmol}\\ M_{\text{S}}=32\text{kg}/\text{kmol}\end{array}$

$\begin{array}{l}{M}_{\text{air}}=29\text{kg}/\text{kmol}\\ M_{{\text{N}}_{\text{2}}}=28\text{kg}/\text{kmol}\end{array}$

Here, molar mass of air is $M_{\text{air}}$.

Substitute $5$ for $m_{\text{ash}}$ in Equation (I).

$\begin{array}{c}{m}_{\text{total}}=100-5\\ =95\text{kg}\end{array}$

Substitute $\text{61}\text{.40}\text{kg}$ for $m_{\text{C}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (II).

$\begin{array}{c}m{f}_{\text{C}}=\frac{\text{61}\text{.40}\text{kg}}{\text{95}\text{kg}}\\ =0.6463\\ =64.63\text{kg}\end{array}$

Substitute $\text{5}\text{.79}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (III).

$\begin{array}{c}m{f}_{{\text{H}}_{\text{2}}}=\frac{\text{5}\text{.79}\text{kg}}{\text{95}\text{kg}}\\ =0.06095\\ =6.095\text{kg}\end{array}$

Substitute $\text{25}\text{.31}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (IV).

$\begin{array}{c}m{f}_{{\text{O}}_{\text{2}}}=\frac{\text{25}\text{.31}\text{kg}}{\text{95}\text{kg}}\\ =0.2664\\ =26.64\text{kg}\end{array}$

Substitute $\text{1}\text{.09}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (V).

$\begin{array}{c}m{f}_{{\text{N}}_{\text{2}}}=\frac{\text{1}\text{.09}\text{kg}}{\text{95}\text{kg}}\\ =0.01147\\ =1.147\text{kg}\end{array}$

Substitute $\text{1}\text{.41}\text{kg}$ for $m_{\text{S}}$ and $\text{95}\text{kg}$ for $m_{\text{total}}$ in Equation (VI).

$\begin{array}{c}m{f}_{\text{S}}=\frac{\text{1}\text{.41}\text{kg}}{\text{95}\text{kg}}\\ =0.01484\\ =1.484\text{kg}\end{array}$

Substitute $\text{64}\text{.63}\text{kg}$ for $m_{\text{C}}$ and $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$ in Equation (VII).

$\begin{array}{c}{N}_{\text{C}}=\frac{\text{64}\text{.63}\text{kg}}{12\text{kg}/\text{kmol}}\\ =5.386\text{kmol}\end{array}$

Substitute $\text{6}\text{.095}\text{kg}$ for $m_{{\text{H}}_{\text{2}}}$ and $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}}$ in Equation (VIII).

$\begin{array}{c}{N}_{{\text{H}}_{\text{2}}}=\frac{\text{6}\text{.095}\text{kg}}{2\text{kg}/\text{kmol}}\\ =3.048\text{kmol}\end{array}$

Substitute $\text{26}\text{.64}\text{kg}$ for $m_{{\text{O}}_{\text{2}}}$ and $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$ in Equation (IX).

$\begin{array}{c}{N}_{{\text{O}}_{\text{2}}}=\frac{\text{26}\text{.64}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.8325\text{kmol}\end{array}$

Substitute $\text{1}\text{.147}\text{kg}$ for $m_{{\text{N}}_{\text{2}}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (X).

$\begin{array}{c}{N}_{{\text{N}}_{\text{2}}}=\frac{\text{1}\text{.147}\text{kg}}{28\text{kg}/\text{kmol}}\\ =0.04096\text{kmol}\end{array}$

Substitute $\text{1}\text{.484}\text{kg}$ for $m_{\text{S}}$ and $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ in Equation (XI).

$\begin{array}{c}{N}_{\text{S}}=\frac{\text{1}\text{.484}\text{kg}}{32\text{kg}/\text{kmol}}\\ =0.04638\text{kmol}\end{array}$

Substitute $5.386\text{kmol}$ for $N_{\text{C}}$, $3.048\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.8325\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $0.04096\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $0.04638\text{kmol}$ for $N_{\text{S}}$ in Equation (XII).

$\begin{array}{c}{N}_m=5.386\text{kmol}+3.048\text{kmol}+0.8325\text{kmol}+0.04096\text{kmol}+0.04638\text{kmol}\\ =9.354\text{kmol}\end{array}$

Substitute $5.386\text{kmol}$ for $N_{\text{C}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XIII).

$\begin{array}{c}{y}_{\text{C}}=\frac{5.386\text{kmol}}{9.354\text{kmol}}\\ =0.5758\end{array}$

Substitute $3.048\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XIV).

$\begin{array}{c}{y}_{{\text{H}}_{\text{2}}}=\frac{3.048\text{kmol}}{9.354\text{kmol}}\\ =0.3258\end{array}$

Substitute $0.8325\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XV).

$\begin{array}{c}{y}_{{\text{O}}_{\text{2}}}=\frac{0.8325\text{kmol}}{9.354\text{kmol}}\\ =0.0890\end{array}$

Substitute $0.04096\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XVI).

$\begin{array}{c}{y}_{{\text{N}}_{\text{2}}}=\frac{0.04096\text{kmol}}{9.354\text{kmol}}\\ =0.00438\end{array}$

Substitute $0.04638\text{kmol}$ for $N_{\text{S}}$ and $9.354\text{kmol}$ for $N_m$ in Equation (XVII).

$\begin{array}{c}{y}_{\text{S}}=\frac{0.04638\text{kmol}}{9.354\text{kmol}}\\ =0.00496\end{array}$

Express the combustion equation.

$\left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +a_{\text{th}}\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to x\left(\text{0}{\text{.95CO}}_{\text{2}}+0.05\text{CO}\right)\\ +y{\text{H}}_{\text{2}}\text{O}+z{\text{SO}}_2+k{\text{N}}_{\text{2}}\end{array}\right]$ (XXVII)

Perform the species balancing:

Carbon balance:

$x=0.5758$

Hydrogen balance:

$y=0.3258$

Sulphur balance:

$z=0.00496$

Oxygen balance:

$\begin{array}{l}0.0890+a_{\text{th}}=0.95x+\left(0.5\times 0.05x\right)+0.5y+z\\ 0.0890+a_{\text{th}}=0.95\left(0.5758\right)+\left(0.5\times 0.05\times 0.5758\right)+0.5\left(0.3258\right)+0.00496\\ 0.0890+a_{\text{th}}=0.72925\\ a_{\text{th}}=0.6403\end{array}$

Nitrogen balance:

$\begin{array}{c}k=0.00438+3.76a_{\text{th}}\\ =0.00438+3.76\left(0.6403\right)\\ =2.412\end{array}$

Substitute $0.5758$ for $x$, $0.3258$ for $y$, $0.00496$ for $z$, $0.6403$ for $a_{\text{th}}$ and $2.412$ for $k$ in Equation (XXVII).

$\left[\begin{array}{l}0.5758\text{C}+0.3258{\text{H}}_{\text{2}}+0.0890{\text{O}}_{\text{2}}+0.00438{\text{N}}_{\text{2}}+0.00496\text{S}\\ +0.6403\left({\text{O}}_{\text{2}}+3.7{\text{6N}}_{\text{2}}\right)\to 0.5470{\text{CO}}_{\text{2}}+0.0288\text{CO}\\ +0.3258{\text{H}}_{\text{2}}\text{O}+0.00496{\text{SO}}_2+2.412{\text{N}}_{\text{2}}\end{array}\right]$ (XXVIII)

Refer Equation (XXVIII), and write the number of moles of products.

$\begin{array}{l}{N}_{{\text{CO}}_2}=0.5470\text{kmol}\\ N_{\text{CO}}=0.0288\text{kmol}\\ N_{{\text{H}}_{\text{2}}\text{O}}=0.3258\text{kmol}\end{array}$

$\begin{array}{l}{N}_{{\text{SO}}_2}=\text{0}\text{.00496}\text{kmol}\\ N_{{\text{N}}_{\text{2}}}=2.412\text{kmol}\end{array}$

Refer Table A-1, “molar mass, gas constant, and the critical point properties”, and write the molar masses.

$\begin{array}{l}{M}_{{\text{CO}}_{\text{2}}}=\text{44}\text{kg}/\text{kmol}\\ M_{\text{CO}}=\text{28}\text{kg}/\text{kmol}\\ M_{{\text{H}}_{\text{2}}\text{O}}=\text{18}\text{kg}/\text{kmol}\end{array}$

$\begin{array}{l}{M}_{{\text{SO}}_{\text{2}}}=\text{64}\text{kg}/\text{kmol}\\ M_{{\text{N}}_{\text{2}}}=\text{28}\text{kg}/\text{kmol}\end{array}$

Substitute $0.5470\text{kmol}$ for $N_{{\text{CO}}_2}$, $0.0288\text{kmol}$ for $N_{\text{CO}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}\text{O}}$, $\text{0}\text{.00496}\text{kmol}$ for $N_{{\text{SO}}_2}$, $2.412\text{kmol}$ for $N_{{\text{N}}_{\text{2}}}$, $\text{44}\text{kg}/\text{kmol}$ for $M_{{\text{CO}}_{\text{2}}}$, $\text{28}\text{kg}/\text{kmol}$ for $M_{\text{CO}}$, $\text{18}\text{kg}/\text{kmol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$, $\text{64}\text{kg}/\text{kmol}$ for $M_{{\text{SO}}_{\text{2}}}$ and $\text{28}\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (XVIII).

$\begin{array}{c}{m}_{\text{total}}=\left[\begin{array}{l}\left(0.5470\text{kmol}\right)\left(\text{44}\text{kg}/\text{kmol}\right)+\left(0.0288\text{kmol}\right)\left(\text{28}\text{kg}/\text{kmol}\right)\\ +\left(0.3258\text{kmol}\right)\left(\text{18}\text{kg}/\text{kmol}\right)+\left(\text{0}\text{.00496}\text{kmol}\right)\left(\text{64}\text{kg}/\text{kmol}\right)+\\ \left(2.412\text{kmol}\right)\left(\text{28}\text{kg}/\text{kmol}\right)\end{array}\right]\\ =98.6\text{kg}\end{array}$

Substitute $0.5470\text{kmol}$ for $N_{{\text{CO}}_2}$, $\text{44}\text{kg}/\text{kmol}$ for $M_{{\text{CO}}_{\text{2}}}$ and $\text{98}\text{.6}\text{kg}$ for $m_{\text{total}}$ in Equation (XIX).

$\begin{array}{c}m{f}_{{\text{CO}}_2}=\frac{\left(0.5470\text{kmol}\right)\left(\text{44}\text{kg}/\text{kmol}\right)}{\text{98}\text{.6}\text{kg}}\\ =0.2441\end{array}$

Hence, the mass fraction of carbon dioxide is $0.2441$.

Substitute $0.0288\text{kmol}$ for $N_{\text{CO}}$, $\text{28}\text{kg}/\text{kmol}$ for $M_{\text{CO}}$ and $\text{98}\text{.6}\text{kg}$ for $m_{\text{total}}$ in Equation (XX).

$\begin{array}{c}m{f}_{\text{CO}}=\frac{\left(0.0288\text{kmol}\right)\left(\text{28}\text{kg}/\text{kmol}\right)}{\text{98}\text{.6}\text{kg}}\\ =0.0082\end{array}$

Hence, the mass fraction of carbon monoxide is $0.0082$.

Substitute $0.3258\text{kmol}$ for $N_{{\text{H}}_2\text{O}}$, $\text{18}\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2\text{O}}$ and $\text{98}\text{.6}\text{kg}$ for $m_{\text{total}}$ in Equation (XXI).

$\begin{array}{c}m{f}_{{\text{H}}_2\text{O}}=\frac{\left(0.3258\text{kmol}\right)\left(\text{18}\text{kg}/\text{kmol}\right)}{\text{98}\text{.6}\text{kg}}\\ =0.0595\end{array}$

Hence, the mass fraction of water is $0.0595$.

Substitute $0.00496\text{kmol}$ for $N_{{\text{SO}}_2}$, $\text{64}\text{kg}/\text{kmol}$ for $M_{{\text{SO}}_2}$ and $\text{98}\text{.6}\text{kg}$ for $m_{\text{total}}$ in Equation (XXII).

$\begin{array}{c}m{f}_{{\text{SO}}_2}=\frac{\left(0.00496\text{kmol}\right)\left(\text{64}\text{kg}/\text{kmol}\right)}{\text{98}\text{.6}\text{kg}}\\ =0.0032\end{array}$

Hence, the mass fraction of sulphur dioxide is $0.0032$.

Substitute $2.412\text{kmol}$ for $N_{{\text{N}}_2}$, $\text{28}\text{kg}/\text{kmol}$ for $M_{{\text{N}}_2}$ and $\text{98}\text{.6}\text{kg}$ for $m_{\text{total}}$ in Equation (XXIII).

$\begin{array}{c}m{f}_{{\text{N}}_2}=\frac{\left(2.412\text{kmol}\right)\left(\text{28}\text{kg}/\text{kmol}\right)}{\text{98}\text{.6}\text{kg}}\\ =0.6849\end{array}$

Hence, the mass fraction of nitrogen is $0.6849$.

Substitute $0.5470\text{kmol}$ for $N_{{\text{CO}}_2}$, $0.0288\text{kmol}$ for $N_{\text{CO}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_2\text{O}}$, $0.00496\text{kmol}$ for $N_{{\text{SO}}_2}$ and $2.412\text{kmol}$ for $N_{{\text{N}}_2}$ in Equation (XXIV).

$\begin{array}{c}{N}_{m,P}=0.5470\text{kmol}+0.0288\text{kmol}+0.3258\text{kmol}+0.00496\text{kmol}+2.412\text{kmol}\\ =3.319\text{kmol}\end{array}$

Substitute $\text{98}\text{.6}\text{kg}$ for $m_{\text{total}}$ and $3.319\text{kmol}$ for $N_{m,P}$ in Equation (XXV).

$\begin{array}{c}{M}_m=\frac{\text{98}\text{.6}\text{kg}}{3.319\text{kmol}}\\ =\text{29}\text{.71}\text{kg}/\text{kmol}\end{array}$

Hence, the apparent molecular weight of the products is $\text{29}\text{.71}\text{kg}/\text{kmol}$.

Since each $\text{kmol}$ of ${\text{O}}_{\text{2}}$ in air is accompanied by $\text{3}\text{.76}\text{kmol}$ of ${\text{N}}_{\text{2}}$, thus

$\begin{array}{c}{N}_{\text{air}}=1\text{kmol}{\text{O}}_{\text{2}}+3.76\text{kmol}{\text{N}}_{\text{2}}\\ =4.76\text{kmol}\text{air}\end{array}$

Refer Equation (XXVIII), and write the number of moles of reactants.

$\begin{array}{l}{N}_{\text{C}}=0.5758\text{kmol}\\ N_{{\text{H}}_{\text{2}}}=0.3258\text{kmol}\\ N_{{\text{O}}_{\text{2}}}=0.0890\text{kmol}\end{array}$

$\begin{array}{l}{N}_{{\text{N}}_2}=\text{0}\text{.00438}\text{kmol}\\ N_{\text{S}}=0.00496\text{kmol}\end{array}$

Substitute $4.76\text{kmol}\text{air}$ for $N_{\text{air}}$, $29\text{kg}/\text{kmol}$ for $M_{\text{air}}$, $0.5758\text{kmol}$ for $N_{\text{C}}$, $0.3258\text{kmol}$ for $N_{{\text{H}}_{\text{2}}}$, $0.0890\text{kmol}$ for $N_{{\text{O}}_{\text{2}}}$, $\text{0}\text{.00438}\text{kmol}$ for $N_{{\text{N}}_2}$, $0.00496\text{kmol}$ for $N_{\text{S}}$, $12\text{kg}/\text{kmol}$ for $M_{\text{C}}$, $2\text{kg}/\text{kmol}$ for $M_{{\text{H}}_2}$, $32\text{kg}/\text{kmol}$ for $M_{{\text{O}}_{\text{2}}}$, $32\text{kg}/\text{kmol}$ for $M_{\text{S}}$ and $28\text{kg}/\text{kmol}$ for $M_{{\text{N}}_{\text{2}}}$ in Equation (XXVI).

$\begin{array}{c}\text{FA}=\frac{0.6403\left(4.76\text{kmol}\text{air}\right)\left(29\text{kg}/\text{kmol}\right)}{\left[\begin{array}{l}\left(0.5758\text{kmol}\right)\left(12\text{kg}/\text{kmol}\right)+\left(0.3258\text{kmol}\right)\left(2\text{kg}/\text{kmol}\right)\\ +\left(0.0890\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)+\left(\text{0}\text{.00438}\text{kmol}\right)\left(28\text{kg}/\text{kmol}\right)\\ +\left(0.00496\text{kmol}\right)\left(32\text{kg}/\text{kmol}\right)\end{array}\right]}\\ =\frac{88.39\text{kg}}{10.69\text{kg}}\\ =\text{8}\text{.27}\text{kg}\text{air}/\text{kg}\text{fuel}\end{array}$

Hence, the mass of air required per unit mass of fuel burned is $\text{8}\text{.27}\text{kg}\text{air}/\text{kg}\text{fuel}$.