Calculating the Volume of Oxygen Required for Propane Combustion: A Comprehensive Guide

Understanding the chemical processes involved in the combustion of hydrocarbons is essential in various fields, including chemistry, chemical engineering, and environmental science. In this article, we will explore the detailed steps required to calculate the volume of oxygen necessary for the complete combustion of 8.8 grams of propane (C3H8). We will break down the process into several clear, reusable steps, making it accessible for readers with varying levels of expertise.

Propane, a versatile hydrocarbon, is widely used for domestic heating, cooking, and industrial applications. To completely burn a given amount of propane, it must react with an appropriate volume of oxygen (O2). This reaction is exothermic, releasing significant amounts of heat and light, as described by the following balanced chemical equation:

The Balanced Chemical Equation for Propane Combustion

The complete combustion of propane (C3H8) is represented by the equation:

C3H8 5O2 → 3CO2 4H2O

This equation tells us that one mole of propane requires five moles of oxygen to undergo complete combustion. This relationship is essential for determining the volume of oxygen needed for a specific amount of propane.

Calculating the Molar Mass of Propane

To accurately calculate the volume of oxygen required, we first need to find the molar mass of propane (C3H8). The molar mass is the sum of the atomic masses of all the atoms in one molecule of the compound.

Carbon (C): 12.01 g/mol Hydrogen (H): 1.008 g/mol

For propane (C3H8):

Carbon: 3 × 12.01 g/mol 36.03 g/mol Hydrogen: 8 × 1.008 g/mol 8.064 g/mol

Therefore, the molar mass of propane is:

Molar mass of C3H8 36.03 g/mol 8.064 g/mol 44.094 g/mol

Converting the Mass of Propane to Moles

The second step involves converting the given mass of propane (8.8 g) into moles. This conversion is necessary to relate the mass to the chemical reaction.

Number of moles of C3H8 can be calculated using the formula:

Moles of C3H8 Mass / Molar Mass 8.8 g / 44.094 g/mol ≈ 0.199 moles

Determining the Moles of Oxygen Needed

Now that we know the moles of propane, we can determine the moles of oxygen required for the complete combustion using the balanced chemical equation. As mentioned earlier, 1 mole of propane reacts with 5 moles of oxygen:

Moles of O2 required Moles of C3H8 × 5 moles of O2/mole of C3H8 ≈ 0.199 moles × 5 ≈ 0.995 moles

Calculating the Volume of Oxygen at STP

The final step is to convert the moles of oxygen into liters, a common unit used in many industrial and scientific contexts. At standard temperature and pressure (STP), 1 mole of an ideal gas occupies 22.4 liters. This is a widely accepted standard used to simplify calculations.

Volume of O2 Moles of O2 × 22.4 L/mol ≈ 0.995 moles × 22.4 L/mol ≈ 22.3 liters

Conclusion

Approximately 22.3 liters of oxygen are required for the complete combustion of 8.8 grams of propane. This process not only helps in understanding the chemical reactions but also in practical applications where the volume of reactants and products is crucial.

Reusable Steps:

Write down the complete equation for the reaction. Delete any redundant steps or calculations for simplicity. Calculate the moles of the reactant using its molar mass. Determine the moles of the necessary product using the balanced equation. Convert the moles of the product into liters using the standard volume at STP.

Keywords: volume of oxygen, propane combustion, chemical equation

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Understanding the importance of chemical equations in combustion processes Exploring the properties and applications of propane in daily life The role of oxygen in various industrial and domestic applications