Understanding Gas Stoichiometry Calculations

Gas stoichiometry is a branch of chemistry that deals with the relationships between the quantities of reactants and products in a chemical reaction, particularly in gaseous states. It often employs the ideal gas law (PV=nRT) to relate pressure, volume, temperature, and the number of moles of gas involved. Below are three diverse, practical examples of gas stoichiometry calculations.

Example 1: Combustion of Propane

In a household setting, propane is commonly used for heating. The combustion of propane is a critical reaction to understand for both safety and efficiency.

The balanced equation for the combustion of propane (C₃H₈) is:

[ C₃H₈ + 5 O₂ \rightarrow 3 CO₂ + 4 H₂O ]

Given: You have 10 moles of propane available for combustion. How many moles of oxygen are required, and what will be the volume of carbon dioxide produced at standard temperature and pressure (STP)?

Calculation:

1. From the balanced equation, 1 mole of propane requires 5 moles of oxygen. Therefore, 10 moles of propane will require:
   [ 10 \text{ moles C₃H₈} \times \frac{5 \text{ moles O₂}}{1 \text{ mole C₃H₈}} = 50 \text{ moles O₂} ]
2. At STP, 1 mole of gas occupies 22.4 liters. Thus, the volume of carbon dioxide produced (3 moles for each mole of propane) is:
   [ 10 \text{ moles C₃H₈} \times 3 \text{ moles CO₂} \times 22.4 \text{ L/mole} = 672 \text{ liters CO₂} ]

Notes: This example illustrates the stoichiometry of combustion reactions and the conversion of moles to volume, which is important in real-world applications like energy production.

Example 2: Production of Ammonia via Haber Process

The Haber process is a method for synthesizing ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) gases, crucial for fertilizer production.

The balanced equation for this reaction is:

[ N₂ + 3 H₂ \rightarrow 2 NH₃ ]

Given: If you start with 8 moles of hydrogen, how many moles of nitrogen are needed, and how many moles of ammonia will be produced?

Calculation:

1. According to the balanced equation, 3 moles of hydrogen require 1 mole of nitrogen. To find the moles of nitrogen needed for 8 moles of hydrogen:
   [ 8 \text{ moles H₂} \times \frac{1 \text{ mole N₂}}{3 \text{ moles H₂}} = \frac{8}{3} \approx 2.67 \text{ moles N₂} ]
2. The moles of ammonia produced will be:
   [ 8 \text{ moles H₂} \times \frac{2 \text{ moles NH₃}}{3 \text{ moles H₂}} = \frac{16}{3} \approx 5.33 \text{ moles NH₃} ]

Notes: Understanding this example is vital for those in agricultural sciences, as ammonia is a key ingredient in fertilizers.

Example 3: Decomposition of Water

Electrolysis is used to decompose water into hydrogen and oxygen gases, a fundamental reaction in renewable energy.

The balanced equation for the electrolysis of water is:

[ 2 H₂O \rightarrow 2 H₂ + O₂ ]

Given: If you start with 20 moles of water, how many moles of hydrogen and oxygen will be produced?

Calculation:

1. From the balanced equation, 2 moles of water yield 2 moles of hydrogen and 1 mole of oxygen. Therefore, from 20 moles of water:

• Moles of hydrogen:
  [ 20 \text{ moles H₂O} \times \frac{2 \text{ moles H₂}}{2 \text{ moles H₂O}} = 20 \text{ moles H₂} ]

• Moles of oxygen:
  [ 20 \text{ moles H₂O} \times \frac{1 \text{ mole O₂}}{2 \text{ moles H₂O}} = 10 \text{ moles O₂} ]

Notes: This example highlights the importance of stoichiometry in energy production, particularly in the context of hydrogen fuel cells.

By understanding these examples of gas stoichiometry calculations, you can apply these concepts to real-world chemical reactions, enhancing both safety and efficiency in various applications.