Question

When ammonium nitrite \(\left(\mathrm{NH}_{4} \mathrm{NO}_{2}\right)\) is heated, it decomposes to give nitrogen gas. This property is used to inflate some tennis balls. (a) Write a balanced equation for the reaction. (b) Calculate the quantity (in grams) of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) needed to inflate a tennis ball to a volume of \(86.2 \mathrm{~mL}\) at \(1.20 \mathrm{~atm}\) and \(22^{\circ} \mathrm{C}\).

Short answer

0.2286 grams of NH4NO2 is needed.

Step-by-step solution

Write the Chemical Equation

The decomposition of ammonium nitrite \( \text{NH}_4\text{NO}_2 \) produces nitrogen gas \( \text{N}_2 \) and water vapor \( \text{H}_2\text{O} \). The balanced chemical equation is:\[\text{NH}_4\text{NO}_2 (s) \rightarrow \text{N}_2 (g) + 2 \text{H}_2\text{O} (g)\]This equation shows that one mole of ammonium nitrite yields one mole of nitrogen and two moles of water vapor.

Apply Ideal Gas Law

Use the ideal gas law to determine the number of moles of nitrogen gas required to fill the tennis ball. The ideal gas law is given by \( PV = nRT \), where:- \( P = 1.20 \text{ atm} \)- \( V = 86.2 \text{ mL} = 0.0862 \text{ L} \)- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \)- \( T = 22^{\circ} \text{C} = 295 \text{ K} \)Solving for \( n \), we have:\[n = \frac{PV}{RT} = \frac{(1.20)(0.0862)}{(0.0821)(295)} \approx 0.00357 \text{ mol}\]This calculation gives the moles of \( \text{N}_2 \) needed.

Calculate Mass of \( \text{NH}_4\text{NO}_2 \)

From the balanced equation, 1 mole of \( \text{NH}_4\text{NO}_2 \) produces 1 mole of \( \text{N}_2 \). Therefore, we need 0.00357 moles of \( \text{NH}_4\text{NO}_2 \). The molar mass of \( \text{NH}_4\text{NO}_2 \) is approximately \( 64.06 \text{ g/mol} \). The mass required is:\[\text{mass} = 0.00357 \text{ mol} \times 64.06 \text{ g/mol} \approx 0.2286 \text{ g}\]So, about 0.2286 grams of ammonium nitrite is needed.

Key concepts

Ammonium Nitrite Decomposition

Ammonium nitrite, represented by the formula \( \text{NH}_4\text{NO}_2 \), is a compound that decomposes upon heating. This process results in the formation of nitrogen gas \( \text{N}_2 \) and water vapor \( \text{H}_2\text{O} \). This decomposition can be illustrated through a balanced chemical equation:

• \( \text{NH}_4\text{NO}_2 (s) \rightarrow \text{N}_2 (g) + 2 \text{H}_2\text{O} (g) \)

This equation signifies that one mole of ammonium nitrite will yield one mole of nitrogen gas and two moles of water vapor. This reaction is significant in applications such as inflating tennis balls, where the gas formed is utilized to achieve the desired volume. Understanding the decomposition of ammonium nitrite is essential for quantifying the reagents needed for practical applications.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics, expressed as \( PV = nRT \). This equation relates the pressure \( P \), volume \( V \), and temperature \( T \) of a gas to the number of moles \( n \) it contains, with \( R \) being the ideal gas constant \( 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \).To apply the Ideal Gas Law, it's important to convert all units to match those used in the equation. For example, if volume is given in milliliters, it should be converted to liters. In a scenario where we need to calculate the amount of nitrogen gas required to inflate a tennis ball to \( 86.2 \text{ mL} \) at \( 1.20 \text{ atm} \) and \( 22^{\circ} \text{C} \) (which is \( 295 \text{ K} \)), we rearrange the equation to solve for \( n \) (moles):\[n = \frac{PV}{RT} = \frac{(1.20)(0.0862)}{(0.0821)(295)} \approx 0.00357 \, \text{mol} \]This calculation gives the amount of nitrogen gas required in moles, showcasing the flexibility and utility of the Ideal Gas Law in predicting how gases behave under different conditions.

Molar Mass Calculation

Molar mass is the mass of one mole of a given compound, measured in grams per mole (g/mol). Calculating molar mass is crucial for converting between grams and moles, especially in stoichiometric calculations.To determine the molar mass of ammonium nitrite \( \text{NH}_4\text{NO}_2 \), you add the atomic masses of each element based on their occurrences in the compound:

• Nitrogen (N): \( 2 \times 14.01 \, \text{g/mol} = 28.02 \, \text{g/mol} \)
• Hydrogen (H): \( 4 \times 1.01 \, \text{g/mol} = 4.04 \, \text{g/mol} \)
• Oxygen (O): \( 2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol} \)

Adding these values gives a molar mass of approximately \( 64.06 \, \text{g/mol} \) for ammonium nitrite. This number is then used to calculate the mass of the compound needed for certain reactions. For instance, to produce \( 0.00357 \, \text{mol} \) of nitrogen gas, you would require \(\text{mass} = 0.00357 \, \text{mol} \times 64.06 \, \text{g/mol} \approx 0.2286 \, \text{g}\).Being able to calculate and understand molar mass is vital for any chemical calculation where mole-based conversions are needed.