Chapter 9, Problem 119QP

Interpretation Introduction

Interpretation:

The mass of ammonium nitrate required to produce $145\text{ L}$ of ${\text{N}}_{\text{2}}\text{O}$ is to be determined.

Concept Introduction:

Ideal gas law is followed by ideal gases. It states that the pressure of gas $\left(P\right)$ and the volume of gas $\left(V\right)$ are directly proportional to the temperature of gas $\left(T\right)$ and the number of moles of gas $\left(n\right)$ .

$PV=nRT$ ……(1)

Answer to Problem 119QP

Solution:

The mass of ammonium nitrate required to produce $145\text{ L}$ of ${\text{N}}_{\text{2}}\text{O}$ is $1.68\times {10}^3\text{ g}$ .

Explanation of Solution

Given Information: The volume of ${\text{N}}_{\text{2}}\text{O}$ is $145\text{ L}$ at $2850\text{ torr}$ and $42°\text{C}$ .

The reaction is as follows:

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\left(s\right)\to {\text{N}}_{\text{2}}\text{O}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The volume of ${\text{N}}_{\text{2}}\text{O}$ is $145\text{ L}$ at $2850\text{ torr}$ and $42°\text{C}$ . The number of moles of ${\text{N}}_{\text{2}}\text{O}$ produced can be calculated using ideal gas law.

Convert degree Celsius temperature in Kelvin:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Temperature $42°\text{C}$ is converted into Kelvin as shown below:

$\begin{array}{c}T=42°\text{C}+273.15\\ =315.2\text{ K}\end{array}$

Convert pressure units from $\text{torr}$ to $\text{atm}$ as shown below:

$\begin{array}{l}1\text{ atm}=760\text{ torr}\\ \text{1 torr}=\frac{1}{760}\text{ atm}\end{array}$

For $2850\text{ torr}$ , pressure in $\text{atm}$ is,

$\begin{array}{c}\text{2850 torr}=\frac{1\text{ atm}}{760\text{ torr}}\times \text{2850 torr}\\ =3.75\text{ atm}\end{array}$

Substitute $P$ as $3.75\text{ atm}$ , $V$ as $145\text{ L}$ , $T$ as $315.2\text{ K}$ and $R$ as $0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}$ in equation (1) to calculate $n$ .

$\begin{array}{c}3.75\text{ atm}\times 145\text{ L}=n\times 0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 315.2\text{ K}\\ n=\frac{3.75\text{ atm}\times 145\text{ L}}{0.08206{\text{ L atm mol}}^{-1}{\text{ K}}^{-1}\times 315.2\text{ K}}\\ =21{\text{ mol N}}_{\text{2}}\text{O}\end{array}$

So, the number of moles of ${\text{N}}_{\text{2}}\text{O}$ formed is $21\text{ mol}$ . According to the reaction, $\text{1 mol}$ of ${\text{N}}_{\text{2}}\text{O}$ gas is produced by $\text{1 mol}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ . Therefore, $21\text{ mol}$ of ${\text{N}}_{\text{2}}\text{O}$ gas is produced by $21\text{ mol}$ of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ .

The mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ $\left(m\right)$ required to form $21\text{ mol}$ of ${\text{N}}_{\text{2}}\text{O}$ gas can be calculated by the product of number of moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ $\left(n\right)$ and molecular mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ $\left(MM\right)$ .

$n_{{\text{NH}}_{\text{4}}{\text{NO}}_3}=\frac{m_{{\text{NH}}_{\text{4}}{\text{NO}}_3}}{M{M}_{{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}}$ …(2)

Substitute $M{M}_{{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}$ as $80{\text{ g mol}}^{-1}$ and $n_{{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}$ as $21\text{ mol}$ in equation (2).

$\begin{array}{c}{m}_{{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}=21\text{ mol}\times 80{\text{ g mol}}^{-1}\\ =1680\text{ g}\\ =1.68\times {10}^3\text{ g}\end{array}$

Conclusion

The mass of ammonium nitrate required to produce $145\text{ L}$ of ${\text{N}}_{\text{2}}\text{O}$ is $1.68\times {10}^3\text{ g}$ .