Question

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) can be produced by thermal decomposition of ammonium nitrate: $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \stackrel{\mathrm{heat}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ What volume of \(\mathrm{N}_{2} \mathrm{O}(g)\) collected over water at a total pressure of 94.0 \(\mathrm{kPa}\) and \(22^{\circ} \mathrm{C}\) can be produced from thermal decomposition of 8.68 \(\mathrm{g} \mathrm{NH}_{4} \mathrm{NO}_{3} ?\) The vapor pressure of water at \(22^{\circ} \mathrm{C}\) is 21 torr.

Short answer

The volume of \(N_2O\) that can be collected over water after thermal decomposition of 8.68 g of \(NH_4NO_3\) at a total pressure of 94.0 kPa and 22°C is approximately 2.843 liters.

Step-by-step solution

Convert mass of \( NH_4NO_3 \) to moles

The molar mass of \( NH_4NO_3 \) is \( 14 + 1\cdot4 + 14 + 16\cdot3 = 80\, g/mol \). To convert the given mass (8.68 g) into moles, divide it by its molar mass: \[ \frac{8.68\,g}{80\, g/mol} = 0.1085\, moles \]

Mole ratio calculation

From the reaction, it can be seen that 1 mole of \( NH_4NO_3 \) will produce 1 mole of \( N_2O \). Hence the amount of \( N_2O \) produced is also 0.1085 moles.

Getting the pressure of \( N_2O \)

The total collected pressure is the sum of the partial pressure of the \( N_2O \) and the water vapor. Given that the vapor pressure of water at \( 22^{\circ}C \) is 21 torr, which is equivalent to 2.8 kPa, the pressure of \( N_2O \) can be found as follows: \[ 94.0\,kPa - 2.8\,kPa = 91.2\,kPa \]

Convert the temperature to Kelvin

To use the ideal gas law, the temperature must be in Kelvin. \( 22^{\circ}C \) is equivalent to \( 22 + 273.15 = 295.15\,K \)

Apply the ideal gas law to find the volume of \( N_2O \)

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is number of moles, \( R \) is the universal gas constant, and \( T \) is temperature. We can rearrange the formula to solve for volume (\( V = \frac{nRT}{P} \)). Substituting the known values: \[ V = \frac{0.1085\,moles \cdot 8.314\,L\cdot kPa/(mol\cdot K) \cdot 295.15\,K}{91.2\,kPa} = 2.843\,L \] So, the volume of \( N_2O \) that can be produced from the decomposition of 8.68 g of \( NH_4NO_3 \) under the given conditions is approximately 2.843 liters.

Key concepts

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that allows us to bridge the gap between the mass of a substance and the amount (in moles) of molecules or atoms it contains. It's calculated by summing up all the atomic masses of the atoms present in a molecule. For example, in ammonium nitrate, or \( \mathrm{NH}_4\mathrm{NO}_3 \), we have:

• Nitrogen (N): 2 atoms at \( 14\, \mathrm{g/mol} \) each
• Hydrogen (H): 4 atoms at \( 1\, \mathrm{g/mol} \) each
• Oxygen (O): 3 atoms at \( 16\, \mathrm{g/mol} \) each

The total calculation becomes \( 14 \times 2 + 1 \times 4 + 16 \times 3 = 80\, \mathrm{g/mol} \). This means for every mole of ammonium nitrate, it weighs 80 grams. This conversion is crucial when you're trying to relate the mass of a substance to its chemical behavior in a reaction, as seen in our exercise where we converted grams to moles.

Vapor Pressure

Vapor pressure is an important concept when dealing with gases, especially when they are collected in conditions involving liquids. It refers to the pressure exerted by the vapor of a liquid (or solid) in the surrounding space when the substance is in equilibrium with its own phase.At \( 22^{\circ} \mathrm{C} \), the vapor pressure for water is noted as 21 torr, which is about 2.8 kPa. This means that at this temperature, the water alone exerts a pressure of 2.8 kPa in the surrounding space. In practice, this value is critical because when you collect a gas over water, you must subtract the water's vapor pressure from the total pressure to determine the pressure of the gas itself. In our problem, the total pressure was 94.0 kPa including the vapor pressure of the water, giving us the pressure of \( \mathrm{N}_2\mathrm{O} \) at 91.2 kPa.

Gas Stoichiometry

Gas stoichiometry involves quantitative relationships between reactants and products in a chemical reaction when they are in the form of gases. It is an application of the principle of stoichiometry combined with the ideal gas law.Using the chemical equation \( \mathrm{NH}_4\mathrm{NO}_3 \rightarrow \mathrm{N}_2\mathrm{O} + 2\mathrm{H}_2\mathrm{O} \), we see that one mole of ammonium nitrate yields one mole of nitrous oxide. This gives us a direct conversion factor for moles to moles, simplifying calculations involving gaseous products.Thanks to the ideal gas law \( PV = nRT \), where:

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the moles of gas,
• \( R \) is the ideal gas constant, and
• \( T \) is the temperature,

we are able to calculate the volume of gas produced. In our scenario, knowing \( n = 0.1085 \) moles and substituting the values, we find that the volume \( V \) for \( \mathrm{N}_2\mathrm{O} \) is about 2.843 liters. This highlights how easily we can relate different physical states and quantities using the principles of gas stoichiometry.