Question

Ammonium nitrite, ${\mathrm{NH}}_4{\mathrm{NO}}_2$, decomposes according to the chemical equation below. $${\mathrm{NH}}_4{\mathrm{NO}}_2(\mathrm{s})⟶{\mathrm{N}}_2(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$$ What is the total volume of products obtained when $128\mathrm{g}{\mathrm{NH}}_4{\mathrm{NO}}_2$ decomposes at ${819}^{\circ }\mathrm{C}$ and $101\mathrm{kPa}?$

Short answer

The total volume of the products obtained when 128 g of ${\mathrm{NH}}_4{\mathrm{NO}}_2$ decomposes under the given conditions is approximately 406.9 L.

Step-by-step solution

Calculate Moles

First calculate the number of moles of ${\mathrm{NH}}_4{\mathrm{NO}}_2$ using the formula $\mathrm{Moles}=\frac{\mathrm{Mass}}{\mathrm{Molar}\mathrm{mass}}$. From the periodic table, the molar mass of ${\mathrm{NH}}_4{\mathrm{NO}}_2$ is approximately $32+14+16\ast 2=78\mathrm{g}/\mathrm{mol}$. Thus, the number of moles of ${\mathrm{NH}}_4{\mathrm{NO}}_2$ is $\frac{128\mathrm{g}}{78\mathrm{g}/\mathrm{mol}}\approx 1.64\mathrm{moles}$.

Determine Moles of Gas Produced

From the stoichiometric coefficients in the balanced chemical reaction, one mole of ${\mathrm{NH}}_4{\mathrm{NO}}_2$ produces one mole of ${\mathrm{N}}_2$ gas and 2 moles of ${\mathrm{H}}_2\mathrm{O}$ gas. Therefore, the total moles of gas produced is $1.64\mathrm{moles}\ast (1+2)=4.92\mathrm{moles}$.

Apply the Ideal Gas Law

The ideal gas law is $\mathrm{PV}=\mathrm{nRT}$, where $\mathrm{P}$ is the pressure in $\mathrm{kPa}$, $\mathrm{V}$ is the volume in $\mathrm{l}$, $\mathrm{n}$ is the moles of gas, $\mathrm{R}$ is the ideal gas constant = $8.31\mathrm{kPa}\cdot \mathrm{L}\cdot {\mathrm{K}}^{-1}\cdot {\mathrm{mol}}^{-1}$, and $\mathrm{T}$ is the Kelvin temperature. First convert the temperature from Celsius to Kelvin by adding 273, giving $\mathrm{T}=819+273=1092\mathrm{K}$. Substituting the given values into the ideal gas law, the volume $\mathrm{V}$ is obtained by $\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\frac{4.92\mathrm{moles}\ast 8.31\mathrm{kPa}\cdot \mathrm{L}\cdot {\mathrm{K}}^{-1}\cdot {\mathrm{mol}}^{-1}\ast 1092\mathrm{K}}{101\mathrm{kPa}}\approx 406.9\mathrm{L}$.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry used to relate the physical properties of gases. It is expressed as $PV=nRT$, where:

• $P$ is the pressure of the gas in kilopascals (kPa)
• $V$ is the volume of the gas in liters (L)
• $n$ is the number of moles of the gas
• $R$ is the ideal gas constant, $8.31\text{kPa}\cdot \text{L}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$
• $T$ is the temperature in Kelvin (K)

To use this law effectively, always make sure the temperature is in Kelvin by adding 273 to the Celsius measurement. In the exercise, converting ${819}^{\circ }\text{C}$ to Kelvin gave $1092\text{K}$.
This law helps us calculate volume when pressure, temperature, and moles of gas are known, as demonstrated in the step-by-step solution.

Stoichiometry

Stoichiometry involves calculations based on the balanced chemical equations. It lets us determine relationships between reactants and products.
In our exercise, the decomposition of ammonium nitrite $({\text{NH}}_4{\text{NO}}_2)$ was examined. The balanced equation shows that each mole of ${\text{NH}}_4{\text{NO}}_2$ yields:

• 1 mole of nitrogen gas $({\text{N}}_2)$
• 2 moles of water vapor $({\text{H}}_2\text{O})$

By knowing the initial amount (moles) of ${\text{NH}}_4{\text{NO}}_2$, we can determine the total moles of gas produced. For this, we used the formula:$$\text{Total Moles of Gas}=(1+2)\times \text{Moles of }{\text{NH}}_4{\text{NO}}_2$$In the example, 1.64 moles of ${\text{NH}}_4{\text{NO}}_2$ decomposed to give 4.92 moles of gas.

Ammonium Nitrite

Ammonium nitrite $({\text{NH}}_4{\text{NO}}_2)$ is a chemical compound that can decompose into nitrogen gas and water vapor. In its solid form, it undergoes a simple breakdown:

• The reaction is given by ${\text{NH}}_4{\text{NO}}_2(\text{s})\to {\text{N}}_2(\text{g})+2{\text{H}}_2\text{O}(\text{g})$

This decomposition involves a rearrangement of the elements, resulting in gas production, which allows us to use the Ideal Gas Law to find total volume.
Ammonium nitrite itself is generally unstable. It is used in controlled settings in labs to understand gas law applications and stoichiometric calculations, similar to the exercise solution. Understanding its decomposition is crucial for practical applications where gas generation is involved.