Ethylene And Oxygen Reaction How Many Grams React At RTP

Jul 10, 2025 by ADMIN 57 views

How Much Ethylene Reacts with 50.5 L of Oxygen at RTP? A Comprehensive Guide

Understanding stoichiometry and chemical reactions is crucial in chemistry. In this comprehensive guide, we will delve into a specific problem involving the reaction of ethylene (C2H4) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). We will break down the problem step by step, ensuring a clear understanding of the concepts involved. This article is designed to help students, educators, and anyone interested in chemistry grasp the fundamentals of stoichiometric calculations and gas reactions at Room Temperature and Pressure (RTP).

Understanding the Chemical Reaction

To accurately determine how much ethylene reacts, it’s essential to first understand the balanced chemical equation for the reaction. The balanced equation provides the molar ratios of reactants and products, which is the foundation for stoichiometric calculations. In this case, the reaction between ethylene (C2H4) and oxygen (O2) produces carbon dioxide (CO2) and water (H2O). The balanced chemical equation is:

C2H4 + 3O2 → 2CO2 + 2H2O

This equation tells us that one molecule of ethylene reacts with three molecules of oxygen to produce two molecules of carbon dioxide and two molecules of water. More importantly, it tells us that one mole of ethylene reacts with three moles of oxygen. This 1:3 molar ratio between ethylene and oxygen is crucial for solving our problem. When dealing with gases, it's also vital to consider the conditions under which the reaction takes place, such as temperature and pressure, as these can affect the volume and amount of gas involved. At Room Temperature and Pressure (RTP), which is typically defined as 20°C (293 K) and 1 atmosphere (101.3 kPa), one mole of any gas occupies approximately 24 liters. This molar volume at RTP is a key piece of information for converting between volume and moles, which is often necessary in stoichiometric calculations involving gases.

Calculating Moles of Oxygen at RTP

The next critical step in determining how much ethylene reacts is to calculate the number of moles of oxygen (O2) involved in the reaction. We are given that 50.5 liters of O2 are reacting at Room Temperature and Pressure (RTP). As mentioned earlier, at RTP, one mole of any gas occupies approximately 24 liters. This provides a direct conversion factor between volume and moles.

To calculate the number of moles of O2, we use the following formula:

Moles of O2 = Volume of O2 / Molar volume at RTP

Plugging in the given values:

Moles of O2 = 50.5 L / 24 L/mol

Moles of O2 ≈ 2.104 moles

Therefore, approximately 2.104 moles of oxygen are reacting. This calculation is fundamental because it allows us to relate the given volume of oxygen to its molar quantity, which is essential for using the stoichiometric ratios from the balanced chemical equation. The molar volume at RTP is a constant that simplifies these calculations, making it easy to convert between the macroscopic property of volume and the microscopic property of moles. Understanding this conversion is crucial for solving many types of stoichiometry problems, especially those involving gases. With the number of moles of oxygen now known, we can proceed to use the balanced chemical equation to determine the corresponding number of moles of ethylene that will react.

Determining Moles of Ethylene Required

With the moles of oxygen (O2) calculated, we can now determine how much ethylene reacts by using the stoichiometric ratio from the balanced chemical equation:

C2H4 + 3O2 → 2CO2 + 2H2O

The balanced equation shows that 1 mole of ethylene (C2H4) reacts with 3 moles of oxygen (O2). This 1:3 ratio is the key to finding the moles of ethylene required. To find the moles of ethylene, we can set up a simple proportion:

(Moles of C2H4) / (Moles of O2) = 1 / 3

We know that the moles of O2 is approximately 2.104 moles, so we can substitute this value into the proportion:

(Moles of C2H4) / 2.104 moles = 1 / 3

To solve for moles of C2H4, we multiply both sides of the equation by 2.104 moles:

Moles of C2H4 = (1 / 3) * 2.104 moles

Moles of C2H4 ≈ 0.7013 moles

Thus, approximately 0.7013 moles of ethylene are required to react completely with 2.104 moles of oxygen. This calculation underscores the importance of the balanced chemical equation in stoichiometry. The coefficients in the balanced equation provide the exact molar ratios needed to ensure that the reaction proceeds correctly, with neither reactant being in excess. Knowing the molar ratio allows us to convert directly from the amount of one substance to the amount of another, which is a fundamental skill in chemistry.

Converting Moles of Ethylene to Grams

After determining the moles of ethylene required for the reaction, the final step in solving how much ethylene reacts is to convert the moles of ethylene (C2H4) to grams. To do this, we need the molar mass of ethylene. The molar mass is the mass of one mole of a substance and is typically expressed in grams per mole (g/mol). Ethylene (C2H4) consists of two carbon atoms and four hydrogen atoms. The atomic mass of carbon (C) is approximately 12.01 g/mol, and the atomic mass of hydrogen (H) is approximately 1.008 g/mol.

To calculate the molar mass of ethylene, we use the following formula:

Molar mass of C2H4 = (2 * Atomic mass of C) + (4 * Atomic mass of H)

Molar mass of C2H4 = (2 * 12.01 g/mol) + (4 * 1.008 g/mol)

Molar mass of C2H4 = 24.02 g/mol + 4.032 g/mol

Molar mass of C2H4 ≈ 28.052 g/mol

Now that we have the molar mass of ethylene, we can convert the moles of ethylene to grams using the formula:

Mass of C2H4 = Moles of C2H4 * Molar mass of C2H4

We previously calculated that approximately 0.7013 moles of ethylene are required. Plugging in the values:

Mass of C2H4 = 0.7013 moles * 28.052 g/mol

Mass of C2H4 ≈ 19.67 grams

Therefore, approximately 19.67 grams of ethylene react with 50.5 L of oxygen at RTP to form carbon dioxide and water. This final calculation ties together all the previous steps, illustrating how the molar mass acts as a bridge between the microscopic world of moles and the macroscopic world of grams. This conversion is essential in chemistry, as it allows us to measure reactants in the lab using grams (a practical unit) and relate those measurements back to the molar quantities that govern chemical reactions.

Conclusion

In conclusion, to determine how much ethylene reacts with 50.5 L of oxygen at RTP, we followed a series of steps grounded in the principles of stoichiometry and gas behavior. We started by understanding the balanced chemical equation, which provided the crucial 1:3 molar ratio between ethylene and oxygen. We then calculated the moles of oxygen using the molar volume at RTP (24 L/mol). By applying the stoichiometric ratio, we found the moles of ethylene required. Finally, we converted the moles of ethylene to grams using its molar mass, arriving at the answer of approximately 19.67 grams of ethylene. This problem-solving process highlights the importance of understanding chemical equations, molar ratios, gas laws, and molar mass in solving quantitative chemistry problems. Mastering these concepts is essential for success in chemistry and related fields.

This step-by-step guide provides a clear framework for approaching similar stoichiometry problems, reinforcing the core concepts and calculations involved. Whether you are a student learning chemistry for the first time or an educator looking for a comprehensive example, this explanation aims to enhance understanding and problem-solving skills in the fascinating world of chemical reactions.