Chapter 14, Problem 39E

Interpretation Introduction

Interpretation:

In the reaction, ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(\text{s}\right)\to {\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)$, the grams of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ exploded when the total volume of gas produced, both dinitrogen oxide and steam, is $\text{82}\text{.3}\text{L}$ at $\text{447}{°}^{}\text{C}$ and $\text{896}\text{torr}$, is to be calculated.

Concept introduction:

Ideal gas is defined as the gas in which the collisions between the molecules and the atoms are perfectly elastic and there are no intermolecular attractive forces found between them. The ideal gas equation is given by the expression as shown below

$\text{PV}=\text{nRT}$

Answer to Problem 39E

The grams of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ exploded when the total volume of gas produced is $\text{82}\text{.3}\text{L}$ at $\text{447}{°}^{}\text{C}$ and $\text{896}\text{torr}$, is $\text{43}\text{.8}\text{g}$.

Explanation of Solution

In the reaction, ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(\text{s}\right)\to {\text{N}}_{\text{2}}\text{O}\left(\text{g}\right)+{\text{2H}}_{\text{2}}\text{O}\left(\text{g}\right)$, the pressure of gases is $\text{896}\text{torr}$ and the temperature is $\text{447}{°}^{}\text{C}$.

The pressure $\left(\text{896}\text{torr}\right)$ can be converted into $\text{atm}$ as shown below.

$\begin{array}{rll}\text{P}& =& \left(896\text{torr}\right)\left(\frac{1\text{atm}}{760\text{torr}}\right)\\ & =& 1.18\text{atm}\end{array}$

The formula to convert the temperature from Celsius to Kelvin is shown below.

$\text{T}\left(\text{K}\right)=\text{T}\left({}^{\text{o}}\text{C}\right)+273$

Therefore, $\text{447}{°}^{}\text{C}$ can be converted to Kelvin as shown below.

$\begin{array}{rll}\text{T}\left(\text{K}\right)& =& \left(\text{447}+273\right)\text{K}\\ & =& 720\text{K}\end{array}$

The temperature of nitrogen gas is $720\text{K}$.

The ideal gas equation is given by the expression as shown below

$\text{PV}=\text{nRT}$

Where,

• $\text{V}$ is the volume occupied by the ideal gas.

• $\text{P}$ is the pressure of the ideal gas.

• $\text{n}$ is the number of moles of the ideal gas.

• $\text{T}$ is the temperature of the ideal gas.

• $\text{R}$ is the ideal gas constant with value $\text{0}\text{.0821}\text{L}\cdot \text{atm/mol}\cdot \text{K}$.

Rearrange the above equation for the value of molar volume $\text{V/n}$.

$\frac{\text{V}}{\text{n}}=\frac{\text{RT}}{\text{P}}$ …(1)

Substitute the values of $\text{P}$, $\text{T}$ and $\text{R}$ in the equation (1).

$\begin{array}{rll}\frac{\text{V}}{\text{n}}& =& \frac{0.0821\frac{\text{L}\cdot \text{atm}}{\text{mol}\cdot \text{K}}\times 720\text{K}}{\text{1}\text{.18}\text{atm}}\\ & =& 50.1\text{L/mol}\end{array}$

Therefore, the molar volume of gases is $50.1\text{L/mol}$.

The volume of gases is $\text{82}\text{.3}\text{L}$.

The formula to convert volume into moles is shown below.

$\text{n}=\frac{\text{V}}{{\text{V}}_{\text{m}}}$ …(2)

Where,

• $\text{V}$ is volume of the substance.

• $\text{n}$ is the number of moles of the substance.

• ${\text{V}}_{\text{m}}$ is the molar volume of the gas.

Substitute the values of volume and molar volume in equation (2).

$\begin{array}{rll}\text{n}& =& \frac{\text{82}\text{.3}\text{L}}{\text{50}\text{.1}\text{L}/\text{mol}}\\ & =& 1.6427\text{mol}\text{gases}\end{array}$

The number of moles of gases that are produced by $\text{1}\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ are $\text{3}\text{mol}$.

Therefore the number of moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left({\text{n}}_{{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}\right)$ that produces $\text{1}\text{.642}\text{mol}\text{gases}$ is calculated as shown below.

$\begin{array}{rll}{\text{n}}_{{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}& =& \frac{\text{1}\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}{\text{3}\text{mol}\text{gases}}\times 1.6427\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\\ & =& 0.5475\text{mol}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\end{array}$

The formula to convert moles into mass is as shown below.

$\text{Mass}=\left(\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\right)$

The molar mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $\text{80}\text{.052}\text{g}/\text{mol}$.

Substitute the values of number of moles and molar mass in the above expression.

$\begin{array}{rll}\text{Mass}& =& \left(\text{0}\text{.5475}\text{mol}\times \text{80}\text{.052}\text{g}/\text{mol}\right)\\ & =& 43.8\text{g}\end{array}$

Therefore, the amount of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ required is $43.8\text{g}$.

Conclusion

The amount of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ exploded when the total volume of gas produced is $\text{82}\text{.3}\text{L}$ at $\text{447}{°}^{}\text{C}$ and $\text{896}\text{torr}$, is $\text{43}\text{.8}\text{g}$.