Determining the Amount of Oxygen Required for the Complete Combustion of Pentane

Understanding the process of complete combustion is crucial for applications ranging from chemical engineering to environmental science. This article will guide you through the calculation of the amount of oxygen required for the complete combustion of 36 grams of pentane (C5H12) under standard conditions, utilizing stoichiometric principles. We will break down the steps to obtain the correct answer, discuss the importance of a balanced chemical equation, and explore the underlying chemistry.

The Stoichiometric Equation for Complete Combustion of Pentane

The complete combustion of pentane can be represented by the following balanced chemical equation:

C5H12(l) 8O2(g) → 5CO2(g) 6H2O(l)

This equation tells us that one mole of pentane requires eight moles of dioxygen (O2) to completely combust, producing five moles of carbon dioxide (CO2) and six moles of water (H2O). The balancing ensures that the number of atoms of each element is the same on both sides of the equation, a fundamental concept in stoichiometry.

Calculating the Moles of Pentane

Let's begin by determining the number of moles of pentane in 36 grams. The molar mass of pentane (C5H12) is calculated as follows:

C: 5 atoms × 12.01 g/mol 60.05 g/mol H: 12 atoms × 1.008 g/mol 12.096 g/mol Total molar mass 60.05 g/mol 12.096 g/mol 72.146 g/mol

Now, we can calculate the moles of pentane in 36 grams:

Moles of pentane 36 g/72.146 g/mol ≈ 0.5 mol

Calculating the Moles of Oxygen Required

According to the balanced equation, each mole of pentane requires 8 moles of O2. Therefore, the moles of O2 required for the combustion of 0.5 moles of pentane can be calculated as:

Moles of O2 required 0.5 mol C5H12 × 8 mol O2/1 mol C5H12 4 mol O2

To convert moles of O2 into grams, we need to know the molar mass of O2. The molar mass of O2 is 32.00 g/mol (since each O2 molecule consists of two oxygen atoms, each with a molar mass of 16.00 g/mol).

Grams of O2 required 4 mol × 32.00 g/mol 128 g

Conclusion and Further Implications

The calculation shows that 128 grams of oxygen are required for the complete combustion of 36 grams of pentane under standard conditions. This knowledge is vital for various applications, such as determining the required air supply in industrial processes or understanding the chemical basis of combustion reactions.

Understanding the stoichiometric relationships between reactants and products in chemical reactions, such as the complete combustion of hydrocarbons like pentane, is fundamental to many fields of science and engineering. This calculation illustrates the importance of accurate stoichiometric analysis and the practical application of chemical principles in real-world scenarios.

Keywords: stoichiometry, complete combustion, oxygen requirements