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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.

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 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.