Mastering Stoichiometry in Advanced Level Chemistry: A Comprehensive Guide
Stoichiometry is a fundamental concept in advanced level chemistry that helps us understand the quantitative relationships between reactants and products in chemical reactions. This article will guide you through solving a complex stoichiometry problem, providing a step-by-step explanation to ensure a thorough understanding of the underlying principles.
Introduction to Stoichiometry
Stoichiometry deals with the relationships between the amounts of reactants and products in a chemical reaction. It involves converting volumes, masses, and amounts of substances into moles using the appropriate chemical formulas and equations. This conversion allows us to determine the theoretical yields, excess reactants, and limiting reactants in a reaction. In this article, we will solve a stoichiometry problem involving sulfur dioxide (SO2) and carbon disulfide (CS2) in an aqueous solution, using the ideal gas law and volume relationships.
Problem Statement
The chemical reaction is as follows:
$$CS_2 3O_2 rightarrow CO_2 2SO_2$$
We are given 100 cm3 of oxygen (O2) and 20 cm3 of carbon disulfide (CS2) in an aqueous solution. The question asks us to determine the volume percentage of the final mixture that consists of the products, specifically SO2 and CO2.
Step-by-Step Solution
Step 1: Determine the Volume of Reactants in Moles
First, we need to convert the given volumes of gases into moles using the ideal gas law. The ideal gas law is given by:
$$PV nRT$$
where:
P is pressure (assuming standard atmospheric pressure of 1 atm), V is volume, T is temperature (assuming standard temperature of 273 K), R is the gas constant (0.0821 L·atm·K-1·mol-1), n is the number of moles.Since the problem doesn’t specify the exact pressure and temperature, we can use the volume directly as it is only important to know the ratio between the volumes of gases.
The ratio of O2 to CS2 in volume is given by:
$$frac{n_{O_2}}{n_{CS_2}} frac{V_{O_2}}{V_{CS_2}} frac{100}{20} 5$$
This means we have 5 times the volume of O2 compared to CS2.
Step 2: Determine the Limiting Reactant
The stoichiometric ratio of CS2 to O2 is 1:3. Therefore, for every 20 cm3 of CS2, we need 60 cm3 of O2 for the reaction to go to completion. However, since we have 100 cm3 of O2, O2 is in excess and CS2 is the limiting reactant.
Step 3: Calculate the Volume of Reactants Used and the Volume of Products Formed
Using the stoichiometry of the reaction:
1 volume of CS2 reacts with 3 volumes of O2 to produce 1 volume of CO2 and 2 volumes of SO2. Therefore, 20 cm3 of CS2 will react with 60 cm3 of O2 to form 20 cm3 of CO2 and 40 cm3 of SO2.Now, the remaining volume of O2 will be:
$$100 - 60 40 text{ cm}^3$$
Step 4: Calculate the Final Volume and Volume Percentage of Products
The final volume of the mixture will be the sum of the unreacted O2 and the products formed:
$$40 text{ cm}^3 20 text{ cm}^3 40 text{ cm}^3 100 text{ cm}^3$$
The volume of the products (CO2 and SO2) is:
$$20 text{ cm}^3 40 text{ cm}^3 60 text{ cm}^3$$
The volume percentage of the products in the final mixture is:
$$frac{60}{100} times 100% 60%$$
Conclusion
This example clearly demonstrates how to approach and solve stoichiometry problems by following a systematic step-by-step process. Understanding the relationships between volumes, moles, and the chemical equations is crucial in advanced level chemistry. By mastering these concepts, students can confidently tackle complex problems and perform accurate calculations.