Calculate The Volume Of Sulfur Trioxide (SO₃) At STP When 0.10 Mol Of O₂(g) Reacts Completely With Excess Sulfur, Given The Reaction: 2 S(s) + 3 O₂(g) → 2 SO₃(g).
Introduction
In the realm of chemistry, stoichiometry plays a pivotal role in understanding the quantitative relationships between reactants and products in chemical reactions. One common application of stoichiometry is calculating the volume of a gas produced in a reaction under specific conditions, such as standard temperature and pressure (STP). This article delves into a problem where 0.10 mol of oxygen gas (O₂) reacts completely with excess sulfur to form sulfur trioxide (SO₃). We will meticulously calculate the volume of SO₃ produced at STP, expressing the final answer in cubic centimeters (cm³). Understanding these calculations is crucial for students and professionals alike in various fields such as chemical engineering, environmental science, and materials science.
Background on Stoichiometry and STP
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Stoichiometric calculations allow us to predict the amounts of reactants and products involved in a chemical reaction.
Standard Temperature and Pressure (STP) is a set of standard conditions for experimental measurements to be established to allow comparisons between different sets of data. According to IUPAC, STP is defined as 0 degrees Celsius (273.15 K) and 100 kPa (1 bar) pressure. At STP, one mole of an ideal gas occupies a volume of 22.7 liters, a value known as the molar volume of a gas at STP. This value is essential for converting between moles and volume under STP conditions.
Problem Statement
Consider the following balanced chemical equation:
2 S(s) + 3 O₂(g) → 2 SO₃(g)
This equation tells us that two moles of solid sulfur (S) react with three moles of gaseous oxygen (O₂) to produce two moles of gaseous sulfur trioxide (SO₃). The problem states that 0.10 moles of O₂(g) react completely with excess sulfur. The task is to calculate the volume of sulfur trioxide (SO₃) that would form at STP, expressed in cm³.
Step-by-Step Solution
To solve this problem, we will follow a step-by-step approach, applying stoichiometric principles and the ideal gas law under STP conditions. By carefully analyzing the balanced chemical equation and using the given information, we can accurately determine the volume of SO₃ produced.
Step 1: Determine the Mole Ratio
The balanced chemical equation provides the stoichiometric relationship between the reactants and products. From the equation:
2 S(s) + 3 O₂(g) → 2 SO₃(g)
We see that 3 moles of O₂ produce 2 moles of SO₃. This mole ratio is crucial for calculating the amount of SO₃ produced from a given amount of O₂.
The mole ratio of O₂ to SO₃ is 3:2. This means that for every 3 moles of O₂ that react, 2 moles of SO₃ are produced. This ratio will be used to calculate the moles of SO₃ produced from 0.10 moles of O₂.
Step 2: Calculate Moles of SO₃ Produced
Using the mole ratio, we can calculate the moles of SO₃ produced from 0.10 moles of O₂:
Moles of SO₃ = (0.10 moles O₂) × (2 moles SO₃ / 3 moles O₂) Moles of SO₃ = (0.10 × 2) / 3 Moles of SO₃ = 0.06666... moles
This calculation shows that 0.10 moles of O₂ will produce approximately 0.0667 moles of SO₃. This value is essential for the next step, where we convert moles of SO₃ to volume at STP.
Step 3: Apply the Molar Volume at STP
At STP, one mole of any ideal gas occupies a volume of 22.7 liters. This molar volume is a standard value used for calculations involving gases at STP. We will use this value to convert the moles of SO₃ calculated in the previous step to the volume in liters.
Volume of SO₃ at STP = Moles of SO₃ × Molar volume at STP Volume of SO₃ at STP = 0.06666... moles × 22.7 L/mole Volume of SO₃ at STP ≈ 1.5133 L
This step converts the moles of SO₃ to liters at STP. The resulting volume is approximately 1.5133 liters, which needs to be converted to cubic centimeters as specified in the problem.
Step 4: Convert Liters to Cubic Centimeters
To convert the volume from liters (L) to cubic centimeters (cm³), we use the conversion factor 1 L = 1000 cm³.
Volume of SO₃ in cm³ = Volume of SO₃ in L × 1000 cm³/L Volume of SO₃ in cm³ = 1.5133 L × 1000 cm³/L Volume of SO₃ in cm³ ≈ 1513.3 cm³
Therefore, the volume of sulfur trioxide (SO₃) formed at STP is approximately 1513.3 cm³.
Answer and Options Analysis
Based on the calculations, the volume of SO₃ formed is approximately 1513.3 cm³. Now, let's analyze the provided options to see which one matches our result:
Options provided: A. (0.10 × 3 × 22.7 × 1000) / 2 B. (0.10 × 3 × 1000) / (2 × ...)
Evaluate option A:
(0.10 × 3 × 22.7 × 1000) / 2 = (0.3 × 22.7 × 1000) / 2 = (6.81 × 1000) / 2 = 6810 / 2 = 3405 cm³
This does not match our calculated volume of approximately 1513.3 cm³.
Let's derive the correct expression based on our calculations:
Moles of SO₃ = (0.10 moles O₂) × (2 moles SO₃ / 3 moles O₂) Moles of SO₃ = (0.10 × 2) / 3
Volume of SO₃ at STP = Moles of SO₃ × 22.7 L/mole × 1000 cm³/L Volume of SO₃ at STP = ((0.10 × 2) / 3) × 22.7 × 1000 cm³ Volume of SO₃ at STP = (0.10 × 2 × 22.7 × 1000) / 3 cm³
Comparing this derived expression with the given options, we can see that none of the provided options exactly match the correct expression. However, if we look closely, we realize that the correct calculation should be:
Volume of SO₃ at STP = (0.10 mol O₂ * (2 mol SO₃ / 3 mol O₂) * 22.7 L/mol * 1000 cm³/L Volume of SO₃ at STP = (0.10 * 2 * 22.7 * 1000) / 3 cm³ Volume of SO₃ at STP ≈ 1513.3 cm³
Based on the given options, it seems there might be a mistake in the options provided, as none of them directly lead to the correct answer. However, the closest option format to the correct formula we derived is:
(0.10 * 2 * 22.7 * 1000) / 3
If there was an option that looked like this, it would be the correct one. Given the options, we must double-check the question and options for any possible errors or misprints.
Conclusion
In this article, we calculated the volume of sulfur trioxide (SO₃) formed at STP from the complete reaction of 0.10 moles of oxygen gas with excess sulfur. The step-by-step solution involved determining the mole ratio from the balanced chemical equation, calculating the moles of SO₃ produced, applying the molar volume at STP, and converting the volume from liters to cubic centimeters. The correct volume of SO₃ formed at STP is approximately 1513.3 cm³. Stoichiometry is fundamental in chemistry, allowing precise calculations of reactant and product quantities in chemical reactions. Mastering these calculations is crucial for success in various scientific and engineering disciplines. Understanding and applying these principles not only solves specific problems but also enhances the ability to analyze and predict chemical behavior in a wide range of contexts.
This exercise highlights the importance of careful stoichiometric calculations and unit conversions in chemistry. While the provided options did not perfectly match the calculated answer, the process of working through the problem demonstrates a clear understanding of the principles involved. Always double-check your calculations and ensure that the units are consistent throughout the problem-solving process. This detailed approach ensures accurate results and enhances your problem-solving skills in chemistry.
For further practice, consider exploring similar stoichiometric problems involving different chemical reactions and conditions. Understanding the principles of stoichiometry is not only crucial for academic success but also for practical applications in various scientific and industrial settings.
By thoroughly understanding the concepts and practicing problem-solving techniques, you can master stoichiometric calculations and confidently tackle more complex chemical problems. This article provides a comprehensive guide to solving such problems, ensuring a solid foundation in stoichiometry.