Problem 84 - Chemistry

Question

Gaseous hydrogen has a density of \(0.0899 \mathrm{~g} / \mathrm{L}\) at \(0{ }^{\circ} \mathrm{C}\), and gaseous chlorine has a density of \(3.214 \mathrm{~g} / \mathrm{L}\) at the same temperature. How many liters of each would you need if you wanted \(1.0078 \mathrm{~g}\) of hydrogen and \(35.45 \mathrm{~g}\) of chlorine?

Step-by-Step Solution

Answer

11.21 L of hydrogen and 11.03 L of chlorine are needed.

Step 1: Understand the Given Values

We are given the densities of hydrogen and chlorine gases, as well as the desired mass for each gas. The density of hydrogen is \(0.0899 \, \text{g/L}\) and that of chlorine is \(3.214 \, \text{g/L}\). We want \(1.0078 \, \text{g}\) of hydrogen and \(35.45 \, \text{g}\) of chlorine.

Step 2: Use Density Formula for Hydrogen

Recall that density \(d\) is defined as \(d = \frac{m}{V}\), where \(m\) is mass and \(V\) is volume. We rearrange to find volume: \(V = \frac{m}{d}\). For hydrogen, plug in the given values: \[ V_H = \frac{1.0078}{0.0899} \approx 11.21 \, \text{L} \].

Step 3: Use Density Formula for Chlorine

Similarly, we use the density formula for chlorine. Substitute the given mass and density of chlorine: \[ V_{Cl} = \frac{35.45}{3.214} \approx 11.03 \, \text{L} \].

Step 4: Assess Results

Now we have the required volumes: \(11.21 \, \text{L}\) of hydrogen and \(11.03 \, \text{L}\) of chlorine. These calculations indicate the amount of each gas needed to reach the given mass based on their densities at the specified conditions.

Key Concepts

Density Formula

The concept of density is fundamental when dealing with states of matter, especially gases. Density is generally defined by the formula:

• \( d = \frac{m}{V} \)

where \( d \) represents density, \( m \) stands for mass, and \( V \) is the volume. This formula helps us understand how much mass a particular volume of a substance contains. In gases, which are less dense than solids and liquids, this formula is crucial for calculating how much space a given mass of gas will occupy.

To find the volume of a gas when its density and mass are known, the density formula can be rearranged to solve for volume:

• \( V = \frac{m}{d} \)

This rearranged formula is particularly useful when you need to determine how much space a certain amount of a gas will take up, as was calculated for hydrogen and chlorine in the exercise. By understanding and utilizing the density formula, one can easily convert between mass and volume for any gaseous substance.

Volume of Gases

When working with gases, it's important to recognize how their volume can change under different conditions. Gases are compressible, meaning their volume can be adjusted by changing pressure or temperature. However, for a given constant temperature and pressure, as in the exercise with hydrogen and chlorine at \(0^\circ C\), their volumes are directly tied to their densities.In the example from the exercise, we calculated the volumes of hydrogen and chlorine needed to reach a specific mass. We used their respective densities to find:

• A volume of approximately \(11.21 \text{ L}\) for hydrogen.
• A volume of approximately \(11.03 \text{ L}\) for chlorine.

As these values demonstrate, even though gases are light and less dense compared to liquids and solids, the volume required to meet a specific mass can be substantial. This reinforces the importance of the density formula in predicting and calculating the spatial requirements of different gases.

Understanding how gases occupy space at the molecular level is critical. Given their low density and high kinetic energy, gas particles are more spread out, resulting in larger volumes for even small masses.

Mass and Volume Relationship

Exploring the relationship between mass and volume helps us comprehend not just gases, but all forms of matter. With gases, as illustrated in the exercise, the connection between mass and volume is mediated by density. Gases make this relationship visibly meaningful due to their large volumes for small masses. For hydrogen and chlorine, their respective masses and densities lead to similar volumes, which is not intuitively obvious without calculation. Here's the interpretation of mass-volume relationship:

• For gases at a constant temperature and pressure, increasing the mass leads to a proportional increase in volume, providing the density remains unchanged.
• Conversely, to maintain a constant mass when density changes, the volume must adjust.

This relationship is crucial in practical applications, such as in chemical reactions where specific volumes of gas are necessary to ensure proper reactant proportions. By knowing the mass and using the density, one can find how much space the gas will occupy, which is critical in lab setups, industrial processes, and environmental studies.