Question

Nitrous oxide (${\text{N}}_{\text{2}}\text{O}$) can be produced by thermal decomposition of ammonium nitrate:

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

What volume of ${\text{N}}_{\text{2}}\text{O(g)}$collected over water at a total pressure of $\text{94.0 kPa}$and${\text{22}}^{\text{o}}\text{C}$can be produced from thermal decomposition of ${\text{8.68 g NH}}_{\text{4}}{\text{NO}}_{\text{3}}$? The vapor pressure of water at${\text{22}}^{\text{o}}\text{C}$is$\text{22torr}$.

Short answer

The volume of ${\text{N}}_{\text{2}}\text{O(g)}$collected over water at a total pressure of $\text{94.0 kPa}$and ${\text{22}}^{\text{o}}\text{C}$is$2.9\cdot {10}^{-3}{m}^3$ which can be produced from thermal decomposition of${\text{8.68 g NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ .

Step-by-step solution

Concept Introduction

The thermal breakdown of organic material at high temperatures and in the absence of oxygen is known as gasification.

The word "thermal breakdown" refers to the process through which heat transforms big molecules into smaller ones.

A chemical species decomposes through a process known as thermal breakdown when its temperature is raised.

Information Provided

The total pressure is:

The temperature in kelvin is:

$$\begin{gathered} \text{p(tot) = 94 kPa} \\ \text{= 94000 Pa} \end{gathered}$$

The vapour pressure of water is: $$\begin{gathered} {\text{t = 22}}^{\text{o}}\text{C} \\ \text{= 273.15 K + 22} \\ \text{= 295.15 K} \end{gathered}$$

The mass of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$is: ${\text{m(NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{) = 8.68 g}$

$$\begin{gathered} {\text{p(H}}_{\text{2}}\text{O) = 22 torr} \\ \text{= 22}·\text{133.322 Pa} \\ \text{= 2800 Pa} \end{gathered}$$

Molar Mass of NH4NO3

At first, calculate the molar mass of compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$because the molar ratio between ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$and ${\text{N}}_{\text{2}}\text{O}$is 1 . This means that for every one reactant molecule product molecule is obtained. Molarity of compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$is the same as molarity of ${\text{N}}_{\text{2}}\text{O}$.

Volume of N2O

Total pressure p(tot)) is the sum of all partial pressures (water vapour pressure and ${\text{N}}_{\text{2}}\text{O}$pressure). Now, calculate the partial pressure of ${\text{N}}_{\text{2}}\text{O}$.

$$\begin{gathered} \text{p(tot) = p}\left({\text{H}}_{\text{2}}\text{O}\right)\text{+ p}\left({\text{N}}_{\text{2}}\text{O}\right) \\ \text{p}\left({\text{N}}_{\text{2}}\text{O}\right)\text{= p(tot) - p}\left({\text{H}}_{\text{2}}\text{O}\right) \\ \text{p}\left({\text{N}}_{\text{2}}\text{O}\right)\text{=94000Pa-2800Pa} \\ \text{p}\left({\text{N}}_{\text{2}}\text{O}\right)\text{=91200Pa} \end{gathered}$$

The last step is to insert all the information in general gas equation to determine the volume of ${\text{N}}_{\text{2}}\text{O}$.

$$\begin{gathered} \text{p}\left({\text{N}}_{\text{2}}\text{O}\right)·\text{V}\left({\text{N}}_{\text{2}}\text{O}\right)\text{= n}\left({\text{N}}_{\text{2}}\text{O}\right)·\text{R}·\text{T} \\ \text{V}\left({\text{N}}_{\text{2}}\text{O}\right)\text{=}\frac{\text{n}\left({\text{N}}_{\text{2}}\text{O}\right)·\text{R}·\text{T}}{\text{p}\left({\text{N}}_{\text{2}}\text{O}\right)} \\ \text{V}\left({\text{N}}_{\text{2}}\text{O}\right)\text{=}\frac{0.108mol.8.314{\displaystyle \frac{Pa.m^3}{mol.K}}295.15K}{91200Pa} \\ \text{V}\left({\text{N}}_{\text{2}}\text{O}\right){\text{=2.9.10}}^{-3}{\text{m}}^3 \end{gathered}$$

Therefore, the value for volume is obtained as$2.9\cdot {10}^{-3}{m}^3.$ .