Question

Given the following amounts of gases, calculate the number of moles of each gas. Calculate the volume each amount of gas would occupy at \(100.0^{\circ} \mathrm{C}\) and \(15.0 \mathrm{~atm}\). (a) \(5.8 \mathrm{~g} \mathrm{NH}_{3}\) (b) \(48 \mathrm{~g} \mathrm{O}_{2}\) (c) \(10.8 \mathrm{gHe}\)

Short answer

The number of moles for (a) NH3 is 0.341 moles, for (b) O2 is 1.5 moles, and for (c) He is 2.70 moles. The volume each gas would occupy at 100.0 C and 15.0 atm is (a) 6.71 L, (b) 30.9 L, and (c) 67.0 L.

Step-by-step solution

Calculate the number of moles

For each gas, divide the given mass by the molar mass: (a) \(NH_3\): \(n = 5.8 \, g / 17.0305 \, g/mol = 0.341 \, mol\), (b) \(O_2\): \(n = 48 \, g / 31.998 \, g/mol = 1.5 \, mol\), (c) \(He\): \(n = 10.8 \, g / 4.0026 \, g/mol = 2.70 \, mol\)

Convert temperature to Kelvin

Subtract 273.15 from the temperature in Celsius to convert to Kelvin: \(T = 100.0 \, ^\circ C + 273.15 = 373.15 \, K\)

Use the Ideal Gas Law to find the volume

Rearrange the Ideal Gas Law to solve for V = nRT/P. Make sure to use the correct value for R, which is 0.08206 L.atm/mol.K. Thus, (a) \(V = 0.341 \, mol \cdot 0.08206 \, L \cdot atm/mol \cdot K \cdot 373.15 \, K / 15.0 \, atm = 6.71 \, L\), (b) \(V = 1.5 \, mol \cdot 0.08206 \, L \cdot atm/mol \cdot K \cdot 373.15 \, K / 15.0\, atm = 30.9 \, L\), (c) \(V = 2.70 \, mol \cdot 0.08206 \, L \cdot atm/mol \cdot K \cdot 373.15 \, K / 15.0 \, atm = 67.0 \, L\)

Key concepts

Ideal Gas Law

The Ideal Gas Law is a critical concept in chemistry, especially when dealing with gases under various conditions of temperature and pressure. It is commonly represented by the equation:
\[ PV = nRT \]
Where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume of the gas,
• \( n \) is the number of moles of the gas,
• \( R \) is the universal gas constant, and
• \( T \) is the temperature of the gas in Kelvins.

This law provides a straightforward method to calculate the volume \( V \) occupied by a gas if the number of moles \( n \), pressure \( P \), and temperature \( T \) are known. The value of the universal gas constant \( R \) depends on the units used for pressure and volume. For example, when pressure is in atmospheres (atm) and volume is in liters (L), the constant \( R \) is typically 0.08206 L.atm/mol.K. Remember, when applying this law, temperature must always be in Kelvins to achieve accurate results.

Molar Mass

Molar mass is a fundamental property in chemistry that relates the mass of a substance to the number of particles it contains. To be precise, it is the mass of one mole of a given substance, where one mole is equal to \(6.022 \times 10^{23}\) entities, known as Avogadro's number.
Calculating the molar mass is essential when converting between mass in grams and moles for any chemical substance. The molar mass can be found by summing the atomic masses of all the atoms in the chemical formula. For instance, for \(NH_3\), the calculation includes one nitrogen (N) atom and three hydrogen (H) atoms. Once you know the molar mass, you can determine the number of moles (\(n\)) by dividing the given mass of the substance by its molar mass, which is the initial step for many stoichiometric calculations.

Gas Volume Conversion

Gas volume conversion is a practical skill when working with gases, as conditions like pressure and temperature can dramatically alter a gas's volume. The Ideal Gas Law enables you to calculate the volume a gas will occupy under new conditions given its original state.
To ensure accuracy in these conversions, it's essential to use consistent units throughout your calculations. Often, volumes must be expressed in liters and pressure in atmospheres to apply the Ideal Gas Law properly. This calculation also necessitates that you convert any temperatures to Kelvin, as the law is dependent on absolute temperature.

Temperature Conversion

Temperature conversion is simple yet crucial when working with gases and applying the Ideal Gas Law. The law requires temperatures to be in Kelvin, which is the SI unit for thermodynamic temperature scale.
To convert from Celsius to Kelvin, which is often necessary in laboratory settings as Celsius is a more commonly encountered temperature scale, you add 273.15 to the Celsius temperature. It's essential not to neglect this conversion, as failing to adjust for the proper temperature scale can result in incorrect calculations when determining the behavior of gases.