Problem 83 - Chemistry The Molecular Science

Question

Mercury(II) azide, \(\mathrm{Hg}\left(\mathrm{N}_{3}\right)_{2},\) is an unstable compound used as a detonator in blasting caps. Calculate the volume (L) of nitrogen produced at \(1 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) when \(2.50 \mathrm{~g}\) mercury azide decomposes to liquid mercury and nitrogen.

Step-by-Step Solution

Answer

Approximately 0.645 L of nitrogen is produced.

Step 1: Write the Balanced Chemical Equation

The decomposition of mercury(II) azide can be represented by the equation: \[ \mathrm{Hg(N_3)_2} \rightarrow \mathrm{Hg} + 3\mathrm{N_2}. \] Here, mercury(II) azide decomposes into liquid mercury and nitrogen gas.

Step 2: Calculate Molar Mass of Hg(N3)2

Find the molar mass of mercury(II) azide (Hg(N3)2). Calculate each component: Mercury (Hg) = 200.59 g/mol, Nitrogen x3 (N3) = 3x14.01 g/mol. Therefore, the total molar mass is \(200.59 + 2\times(3\times14.01) = 284.66\) g/mol.

Step 3: Convert Grams to Moles

Using the molar mass calculated, convert 2.50 g of mercury(II) azide to moles: \[ \text{Moles of } \mathrm{Hg(N_3)_2} = \frac{2.50\, \text{g}}{284.66\, \text{g/mol}} = 0.00878\, \text{mol}. \]

Step 4: Calculate Moles of N2 Produced

According to the balanced equation, 1 mole of Hg(N3)2 produces 3 moles of N2. Calculate moles of N2: \[ 0.00878\, \text{mol of } \mathrm{Hg(N_3)_2} \times \frac{3\, \text{mol } \mathrm{N_2}}{1\, \text{mol } \mathrm{Hg(N_3)_2}} = 0.02634\, \text{mol of } \mathrm{N_2}. \]

Step 5: Use Ideal Gas Law to Find Volume of N2

Use the ideal gas law \( PV = nRT \) to find the volume of \( \mathrm{N_2} \). The conditions are 1 atm (P) and 298.15 K (T = 25°C + 273.15). The ideal gas constant \( R = 0.0821 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \). Calculate: \[ V = \frac{nRT}{P} = \frac{0.02634\, \text{mol} \times 0.0821\, \text{L atm K}^{-1}\,\text{mol}^{-1} \times 298.15\, \text{K}}{1\, \text{atm}} \approx 0.645\, \text{L}. \]

Key Concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It provides a way to calculate how much of each substance is needed or produced in a chemical reaction.

In the decomposition of mercury(II) azide, stoichiometry plays a crucial role. We start by writing the balanced chemical equation, which is the foundation for any stoichiometric calculations:

• Decomposition: \( \mathrm{Hg(N_3)_2} \rightarrow \mathrm{Hg} + 3\mathrm{N_2} \)

The balanced equation tells us that one mole of mercury(II) azide decomposes to produce one mole of liquid mercury and three moles of nitrogen gas (\( \mathrm{N_2} \)). This relationship is essential for our calculations.

Next, we calculate the moles of mercury(II) azide by using its molar mass. Once we have the moles of the reactant, we use the stoichiometric coefficients from the balanced equation to find the moles of nitrogen produced. This step-by-step process ensures accuracy when predicting the amounts of products formed in any given chemical reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. In the case of mercury(II) azide, the reaction is a decomposition, where a complex molecule breaks down into simpler substances.

The chemical equation for the decomposition is:

• \( \mathrm{Hg(N_3)_2} \rightarrow \mathrm{Hg} + 3\mathrm{N_2} \)

Here, the mercury(II) azide decomposes into elemental mercury and nitrogen gas. Understanding this equation is important because it shows the chemical change taking place, tracking all atoms from reactants to products.

This type of decomposition is also exothermic. It can release significant energy, which is why mercury(II) azide is known for its use in detonators. This reaction highlights how chemical reactions can also have practical applications beyond a simple transformation of substances.

Gas Laws

Gas laws describe how gases behave under different conditions of pressure, temperature, and volume. The Ideal Gas Law is a key concept in understanding these relationships, particularly in reactions involving gases.

The Ideal Gas Law is expressed as:

• \( PV = nRT \)
• Where:
• \( P \) = Pressure
• \( V \) = Volume
• \( n \) = Moles of gas
• \( R \) = Ideal gas constant (0.0821 L·atm/K·mol)
• \( T \) = Temperature in Kelvin

In this exercise, we use the Ideal Gas Law to determine the volume of nitrogen gas produced when mercury(II) azide decomposes. By knowing the moles of \( \mathrm{N_2} \), the pressure, and the temperature, we can calculate the volume of gas produced.

This calculation underscores the utility of the Ideal Gas Law in converting between these units, allowing chemists to predict the behavior of gases in various chemical processes.