Question

According to your calculations, a reaction should yield \(5.67 \mathrm{~g}\) of oxygen, \(\mathrm{O}_{2} .\) What do you expect the volume to be at \(23^{\circ} \mathrm{C}\) and 0.894 atm?

Short answer

The expected volume is approximately 4.804 L.

Step-by-step solution

Understand the Problem

The problem asks us to find the volume of oxygen gas at a specific temperature and pressure. We know the mass of the gas which is 5.67 grams, and we are given the temperature as 23°C and pressure as 0.894 atm.

Convert Temperature to Kelvin

The volume of gas can be found using the Ideal Gas Law, but first, we need to convert the temperature from Celsius to Kelvin. The formula for conversion is: \[ T(K) = T(°C) + 273.15 \] So, \[ T = 23 + 273.15 = 296.15 \text{ K} \]

Calculate Moles of Oxygen

The molar mass of oxygen (O₂) is 32.00 g/mol. To find the number of moles of oxygen, use the formula: \[ \text{Moles of } O_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.67 \text{ g}}{32.00 \text{ g/mol}} \approx 0.177 \text{ moles} \]

Use the Ideal Gas Law

Now, apply the Ideal Gas Law, which is given by \[ PV = nRT \] Where:- \( P \) is the pressure in atm- \( V \) is the volume in liters- \( n \) is the number of moles- \( R \) is the ideal gas constant: 0.0821 atm L/mol K- \( T \) is the temperature in Kelvin. Substitute the known values into the equation to solve for volume \( V \): \[ (0.894)V = (0.177)(0.0821)(296.15) \]

Solve for Volume

Calculate the value on the right-hand side:\[ (0.177 \times 0.0821 \times 296.15) \approx 4.297 \]Now solve for \( V \):\[ V = \frac{4.297}{0.894} \approx 4.804 \text{ L} \]

Summary

The volume of oxygen gas at 23°C and 0.894 atm is approximately 4.804 liters.

Key concepts

Temperature Conversion to Kelvin

When working with gas laws, particularly the Ideal Gas Law, always ensure that the temperature is in Kelvin. Kelvin is the absolute temperature scale used in scientific calculations. Converting temperature from Celsius to Kelvin is easy. You just need to add 273.15 to the Celsius temperature.
For example, if you have a temperature of 23°C (like in our exercise), you would convert it to Kelvin as follows:

• Take the Celsius temperature: 23°C.
• Add 273.15 to this temperature.
• The Kelvin temperature becomes: 23°C + 273.15 = 296.15 K.

Using Kelvin ensures that you are working with a scale that starts at absolute zero, which is essential for proper gas law calculations. Always remember: Kelvin = Celsius + 273.15.

Molar Mass of Oxygen

Understanding the molar mass of a molecule is crucial when performing calculations related to substances. The molar mass is the mass of one mole of a substance, and for oxygen gas (O₂), it's 32.00 g/mol. This is because oxygen O₂ is composed of two oxygen atoms.
Each oxygen atom has an atomic mass of about 16.00 g/mol. So, for O₂, you multiply:

• 16.00 g/mol (mass of one oxygen atom) × 2 (since O₂ has two oxygen atoms).
• This gives you: 16.00 g/mol × 2 = 32.00 g/mol for O₂.

Knowing the molar mass helps you convert grams of a substance to moles, which is necessary for using equations like the Ideal Gas Law.

Calculating Moles of Gas

To solve gas law problems, you often need to calculate the number of moles of a gas. Moles are a measure of quantity in chemistry that relates to the number of particles or atoms. To find the moles from a given mass, divide the mass of the substance by its molar mass.
For example, if you have 5.67 grams of oxygen (O₂), you calculate how many moles this is as follows:

• Find the molar mass of oxygen: 32.00 g/mol (from the previous section).
• Divide the mass you have by the molar mass: \[ ext{Moles of } O_2 = \frac{5.67 ext{ g}}{32.00 ext{ g/mol}} \approx 0.177 ext{ moles}. \]

This tells you that 5.67 grams of O₂ is approximately 0.177 moles. Knowing the moles is important because it's used in the Ideal Gas Law equation, which helps you find other properties like volume.