Chapter 6, Problem 6.175AP

Interpretation Introduction

Interpretation: Wetlands are given to be used to treat agricultural waste from rain runoff and groundwater by the technique denitrification which converts nitrate ions to nitrogen gas. The volume of ${\text{N}}_{\text{2}}$ produced at $\text{17}\text{°C}$ and $1\text{atm}$ if the denitrification process is complete and the volume of ${\text{CO}}_{\text{2}}$ produce is to be calculated. If the total pressure is assumed to be $1\text{atm}$ then the density of the mixture is to be calculated.

Concept introduction: The Charles law states that “the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant”.

To determine: The volume of ${\text{N}}_{\text{2}}\text{and}{\text{CO}}_{\text{2}}$ produced and the density of the mixture if the ${\text{P}}_{\text{Total}}$ of the system is assumed to be $1\text{atm}$.

Answer to Problem 6.175AP

Solution

The volume of ${\text{N}}_{\text{2}}\text{and}{\text{CO}}_{\text{2}}$ produced is $\underset{\_}{38.38L}$ and $\underset{\_}{191.84L}$ respectively. The density of mixture is $\underset{\_}{1.73g/L}$.

Explanation of Solution

The given equation is,

$2{\text{NO}}^{-}{}_3(\text{g})+5\text{CO}\text{(g)}+2{\text{H}}^{+}\left(\text{aq}\right)\to {\text{N}}_2\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+5{\text{CO}}_2\left(\text{g}\right)$

Here, 200 g of ${\text{NO}}_{\text{3}}$ are reacting.

The number of moles of ${\text{NO}}_{\text{3}}{}^{-}$ is calculated by the formula,

$n=\frac{\text{Mass}\text{in}\text{grams}}{\text{Molar}\text{mass}\text{in}\text{grams}}$

Substitute the given values of mass and molar mass in the above expression.

$\begin{array}{c}n=\frac{200}{62}\\ =3.225\text{mol}\end{array}$

Molar mass of ${\text{NO}}_{\text{3}}{}^{-}$ is $3.225\text{g/mol}$.

Since the molecular mass $2\times 3.225\text{g/mol}$ of ${\text{NO}}_{\text{3}}{}^{-}$ produces $1\text{mol}\text{of}{\text{N}}_{\text{2}}$.

Therefore $\text{2}\text{mol}$ of ${\text{NO}}_{\text{3}}{}^{-}$ produces $1$ mole of ${\text{N}}_{\text{2}}$ $\begin{array}{c}=\frac{1\times 3.225}{2}\\ =1.6125\text{mol}\text{of}{\text{N}}_{\text{2}}\end{array}$

Thus, the amount of ${\text{N}}_{\text{2}}$ produced will be $1.6125\text{mol}$.

The one mole of any gas occupies $22.4\text{L}$ at STP.

So, the volume occupied by $1.6125\text{mol}\text{of}{\text{N}}_{\text{2}}$ at STP $\begin{array}{c}=\frac{22.4}{1}\times 1.6125\\ =36.12\text{L}\end{array}$

The volume of the ${\text{N}}_{\text{2}}$ gas is calculate by the formula,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where,

• $T_1$ is the initial temperature $290\text{K}$.
• $T_2$ is the new final temperature $17°\text{C}$.
• $V_{\text{1}}$ is the initial volume $36.12\text{L}$.
• $V_2$ is the new final volume.

The conversion of temperature from $°\text{C}\text{to}\text{K}$ is done as,

$K=°C+273.15$

For $T_2$

$\begin{array}{c}{T}_2=17°\text{C}+\text{273}\text{.15}\\ =\text{290}\text{.15}\text{K}\\ \approx 290\text{K}\end{array}$

Substitute these values of $T_1$, $T_2$ and $V_{\text{1}}$ in above expression.

$\begin{array}{c}\frac{V_1}{T_1}=\frac{V_2}{T_2}\\ V_2=\frac{36.12\text{L}\times 290\text{K}}{273\text{K}}\\ =\underset{\_}{38.38L}\end{array}$

Therefore, the volume of ${\text{N}}_{\text{2}}$ at $17°\text{C}$ is $\underset{\_}{38.38L}$.

$2\text{mol}\text{of}{\text{NO}}_{\text{3}}{}^{-}\text{produces}5\text{mol}\text{of}{\text{CO}}_{\text{2}}\text{gas}.$

Molar mass of ${\text{NO}}_{\text{3}}{}^{-}$ is $3.225\text{g/mol}$.

Since the molecular mass $2\times 3.225\text{g/mol}$ of ${\text{NO}}_{\text{3}}{}^{-}$ produces $\text{5}\text{mol}\text{of}{\text{CO}}_{\text{2}}$.

Therefore $\text{2}\text{mol}$ of ${\text{NO}}_{\text{3}}{}^{-}$ produces $5$ mole of ${\text{CO}}_{\text{2}}$ $\begin{array}{c}=\frac{5\times 3.225}{2}\\ =8.0625\text{mol}\text{of}{\text{CO}}_{\text{2}}\end{array}$

Thus, the amount of ${\text{CO}}_{\text{2}}$ produced will be $8.0625\text{mol}$.

The one mole of any gas occupies $22.4\text{L}$ at STP.

So, the volume occupied by $8.0625\text{mol}\text{of}{\text{CO}}_{\text{2}}$ at STP $\begin{array}{c}=\frac{22.4}{1}\times 8.0625\\ =180.6\text{L}\end{array}$

The volume of the ${\text{CO}}_{\text{2}}$ gas is calculated by the formula,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where,

• $T_1$ is the initial temperature $290\text{K}$.
• $T_2$ is the new final temperature $17°\text{C}$.
• $V_{\text{1}}$ is the initial volume $180.6\text{L}$.
• $V_2$ is the new final volume.

The conversion of temperature from $°\text{C}\text{to}\text{K}$ is done as,

$K=°C+273.15$

For $T_2$

$\begin{array}{c}{T}_2=17°\text{C}+\text{273}\text{.15}\\ =\text{290}\text{.15}\text{K}\\ \approx 290\text{K}\end{array}$

Substitute these values of $T_1$, $T_2$ and $V_{\text{1}}$ in above expression.

$\begin{array}{c}\frac{V_1}{T_1}=\frac{V_2}{T_2}\\ V_2=\frac{180.6\text{L}\times 290\text{K}}{273\text{K}}\\ =\underset{\_}{191.84L}\end{array}$

Therefore, the volume of ${\text{CO}}_{\text{2}}$ at $17°\text{C}$ is $\underset{\_}{191.84L}$.

The density of ${\text{CO}}_{\text{2}}$ at STP is $1.96\text{g/L}$ and the density of ${\text{N}}_{\text{2}}$ at STP is $1.255\text{g/L}$.

The density of ${\text{N}}_{\text{2}}$ at $17°\text{C}$ is calculated by the formula,

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

Where,

$T_1$ is the initial temperature.

$T_2$ is the new final temperature.

$V_{\text{1}}$ is the initial volume.

$V_2$ is the new final volume.

Simplify the above expression.

$\frac{\frac{V_1}{M_1}}{\frac{V_2}{M_2}}=\frac{T_1}{T_2}$ (1)

The density is given as,

$\frac{M}{V}=D$

Where,

• $M$ is the molar mass.
• $V$ is the volume.
• $D$ is the density.

Substitute the above relation of density in equation (1).

$\begin{array}{c}\frac{D_1}{D_2}=\frac{T_1}{T_2}\\ D_1=D_2\frac{T_1}{T_2}\end{array}$

Now, substitute the given values in the above expression,

$D_1=D_2\frac{T_1}{T_2}$

Substitute the values of $D_2$, $T_1$ and $T_2$ in the above expression.

$\begin{array}{c}{D}_1=1.255\text{g/L}\times \frac{273\text{K}}{290\text{K}}\\ =1.181\text{g/L}\end{array}$

Similarly, the density of ${\text{CO}}_{\text{2}}$ at $17°\text{C}$ is calculated by the formula,

$\frac{V_1}{V_2}=\frac{T_1}{T_2}$

Where,

• $T_1$ is the initial temperature.
• $T_2$ is the new final temperature.
• $V_{\text{1}}$ is the initial volume.
• $V_2$ is the new final volume.

Simplify the above expression

$\frac{\frac{V_1}{M_1}}{\frac{V_2}{M_2}}=\frac{T_1}{T_2}$

The density is given as,

$\frac{M}{V}=D$

Where,

• $M$ is the molar mass.
• $V$ is the volume.
• $D$ is the density.

Substitute the above relation of density in equation (1).

$\begin{array}{c}\frac{D_1}{D_2}=\frac{T_1}{T_2}\\ D_1=D_2\frac{T_1}{T_2}\end{array}$

Now, substitute the given values in the above expression,

$D_1=D_2\frac{T_1}{T_2}$

Substitute the values of $D_2$, $T_1$ and $T_2$ in the above expression.

$\begin{array}{c}{D}_1=1.96\text{g/L}\times \frac{273\text{K}}{290\text{K}}\\ =1.849\text{g/L}\end{array}$

As gas mixture contains five parts of ${\text{CO}}_{\text{2}}$ and one part of nitrogen. So the density of mixture is calculated by the formula,

$\text{Density}\text{of}\text{mixture}=\frac{5\times \text{Density}\text{of}{\text{CO}}_2+1\times \text{Density}\text{of}{\text{N}}_{\text{2}}}{6}$

Substitute the values of densities of ${\text{CO}}_2$ and ${\text{N}}_{\text{2}}$ in the above expression.

$\begin{array}{c}\text{Density}\text{of}\text{mixture}=\frac{5\times 1.84+1\times 1.181}{6}\\ =\underset{\_}{1.73g/L}\end{array}$

Therefore, density of mixture $\underset{\_}{1.73g/L}$.

Conclusion:

The volume of ${\text{N}}_{\text{2}}\text{and}{\text{CO}}_{\text{2}}$ produced is $\underset{\_}{38.38L}$ and $\underset{\_}{191.84L}$ respectively. The density of mixture is $\underset{\_}{1.73g/L}$