Chapter 6, Problem 6.6P

To determine

The theoretical amount of methane gas produced by anaerobic decomposition.

Answer to Problem 6.6P

The amount of methane gas produced by anaerobic decomposition is $\text{9}\text{.35}\times {\text{10}}^8\text{L}$.

Explanation of Solution

Given:

Amount of material placed in landfill is $10000\text{metric}\text{tonnes}$ .

Calculation:

Calculate the amount of glucose.

  $A_{\text{glucose}}=A_{\text{landfill}}\times 0.2$    .........(I)

Here, the amount of glucose in the land fill is $A_{\text{glucose}}$ and the amount of landfill is $A_{\text{landfill}}$ .

Substitute $10000\text{tonnes}$ for $A_{\text{landfill}}$ in Equation (I).

  $\begin{array}{c}{A}_{\text{glucose}}=\left(10000\text{tonnes}\right)\times 0.2\\ =2000\text{tonnes}\end{array}$

Write the equation for the methane production from glucose.

  $\left(\begin{array}{l}{\text{C}}_a{\text{H}}_b{\text{O}}_c{\text{N}}_d+\left(\frac{4a-b-2c-3d}{4}\right){\text{H}}_{\text{2}}\text{O}\to \\ \left(\frac{4a+b-2c-3d}{8}\right){\text{CH}}_4+\left(\frac{4a-b+2c+3d}{8}\right){\text{CO}}_2+d{\text{NH}}_3\end{array}\right)$ ...... (II)

Here, the stoichiometric constants are $a$ , $b$ , $c$ and $d$ .

The organic considered is glucose whose chemical formula is ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}$ .

Calculate the values of $a$ , $b$ , $c$ and $d$ .

  $\begin{array}{l}a=6\\ b=12\\ c=6\\ d=0\end{array}$

Substitute $6$ for $a$ , $12$ for $b$ , $6$ for $c$ and $0$ for $d$ in Equation (II).

  $\begin{array}{l}\left(\begin{array}{l}{\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{N}}_0+\left(\frac{\left(4\times 6\right)-\left(12\right)-\left(2\times 6\right)}{4}\right){\text{H}}_{\text{2}}\text{O}\to \\ \left(\frac{\left(4\times 6\right)+12-\left(2\times 6\right)}{8}\right){\text{CH}}_4+\left(\frac{\left(4\times 6\right)-12+\left(2\times 6\right)}{8}\right){\text{CO}}_2\end{array}\right)\\ \left({\text{C}}_6{\text{H}}_{12}{\text{O}}_6+\left(0\right){\text{H}}_{\text{2}}\text{O}\to \left(3\right){\text{CH}}_4+\left(3\right){\text{CO}}_2\right)\end{array}$ ...... (III)

Write the expression to calculate the moles of methane produced from glucose.

  $n=\frac{A_{\text{glucose}}}{M_{\text{glucose}}}$     .........(IV)

Here, the moles of methane produced from glucose is $n$ and the molecular mass of glucose is $M_{\text{glucose}}$ .

Substitute $2000\text{tonnes}$ for $A_{\text{glucose}}$ and $180\text{g}/\text{mol}$ for $M_{\text{glucose}}$ in Equation (IV).

  $\begin{array}{c}n=\frac{\left(2000\text{tonnes}\right)\left(\frac{{10}^6\text{g}}{\text{1}\text{ton}}\right)}{180\text{g}/\text{mol}}\\ =1.1\times {10}^7\text{mol}\end{array}$

Calculate the amount of cellulose.

  $A_{\text{cellulose}}=A_{\text{landfill}}\times 0.5$    .........(V)

Here, the amount of cellulose in the land fill is $A_{\text{cellulose}}$ and the amount of landfill is $A_{\text{landfill}}$ .

Substitute $10000\text{tonnes}$ for $A_{\text{landfill}}$ in Equation (V).

  $\begin{array}{c}{A}_{\text{cellulose}}=\left(10000\text{tonnes}\right)\times 0.5\\ =5000\text{tonnes}\end{array}$

Write the equation for the methane production from cellulose.

  $\left(\begin{array}{l}{\text{C}}_a{\text{H}}_b{\text{O}}_c{\text{N}}_d+\left(\frac{4a-b-2c-3d}{4}\right){\text{H}}_{\text{2}}\text{O}\to \\ \left(\frac{4a+b-2c-3d}{8}\right){\text{CH}}_4+\left(\frac{4a-b+2c+3d}{8}\right){\text{CO}}_2+d{\text{NH}}_3\end{array}\right)$ ...... (VI)

Since, the organic considered is cellulose whose chemical formula is ${\text{C}}_{\text{6}}{\text{H}}_{\text{10}}{\text{O}}_{\text{5}}$ .

Calculate the values of $a,b,c$ and $d$ .

  $\begin{array}{l}a=6\\ b=10\\ c=5\\ d=0\end{array}$

Substitute $6$ for $a$ , $10$ for $b$ , $5$ for $c$ and $0$ for $d$ in Equation (VI).

  $\begin{array}{l}\left(\begin{array}{l}{\text{C}}_6{\text{H}}_{10}{\text{O}}_5{\text{N}}_0+\left(\frac{\left(4\times 6\right)-\left(10\right)-\left(2\times 5\right)}{4}\right){\text{H}}_{\text{2}}\text{O}\to \\ \left(\frac{\left(4\times 6\right)+10-\left(2\times 5\right)}{8}\right){\text{CH}}_4+\left(\frac{\left(4\times 6\right)-10+\left(2\times 5\right)}{8}\right){\text{CO}}_2\end{array}\right)\\ \left({\text{C}}_6{\text{H}}_{10}{\text{O}}_5+\left(1\right){\text{H}}_{\text{2}}\text{O}\to \left(3\right){\text{CH}}_4+\left(3\right){\text{CO}}_2\right)\end{array}$

Write the expression to calculate the moles of methane produced from cellulose.

  $n=\frac{A_{\text{cellulose}}}{M_{\text{cellulose}}}$ ...... (VII)

Here, the moles of methane produced from cellulose is $n$ and the molecular mass of cellulose is $M_{\text{cellulose}}$ .

Substitute $5000\text{tonnes}$ for $A_{\text{cellulose}}$ and $162\text{g}/\text{mol}$ for $M_{\text{cellulose}}$ in Equation (VII).

  $\begin{array}{c}n=\frac{5000\text{tonnes}}{162\text{g}/\text{mol}}\times \left(\frac{{10}^{6}g}{\text{tonnes}}\right)\\ =3.08\times {10}^7\text{mol}\end{array}$

Write the expression to calculate the methane produced from cellulose.

  $V=n\times \frac{22.4\text{L}}{1\text{mole}}$ ...... (VIII)

Here, the number of moles of methane produced from cellulose is $n$ and the volume of methane produced is from cellulose is $V_{\text{C}}$ .

Substitute $3.08\times {10}^7\text{mol}$ for $n$ in Equation (VIII).

  $\begin{array}{c}{V}_{\text{C}}=3.08\times {10}^7\text{mol}\times \frac{22.4\text{L}}{1\text{mole}}\\ =6.89\times {10}^8\text{L}\end{array}$

Write the expression to calculate the methane produced from glucose.

  $V_{\text{G}}=n\times \frac{22.4\text{L}}{1\text{mole}}$    .........(IX)

Here, the number of moles of methane produced from glucose is $n$ and the volume of methane produced is from glucose is $V_{\text{G}}$ .

Substitute $1.1\times {10}^7\text{mol}$ for $n$ in Equation (IX).

  $\begin{array}{c}{V}_{\text{G}}=1.1\times {10}^7\text{mol}\times \frac{22.4\text{L}}{1\text{mole}}\\ =2.46\times {10}^8\text{L}\end{array}$

Calculate the total methane produced.

  $V_{\text{T}}=V_{\text{C}}+V_{\text{G}}$     .........(X)

Here, the total theoretical methane produced is $V_{\text{T}}$ .

Substitute $2.46\times {10}^8\text{L}$ for $V_{\text{G}}$ and $6.89\times {10}^8\text{L}$ for $V_{\text{T}}$ in Equation (X).

  $\begin{array}{c}{V}_{\text{T}}=2.46\times {10}^8\text{L}+6.89\times {10}^8\text{L}\\ \text{=9}\text{.35}\times {\text{10}}^8\text{L}\end{array}$

Conclusion:

Thus, the total theoretical methane produced is $V_{\text{T}}$ which is assumed to find the answer.