Calculating Carbon Dioxide Volume From Ethene Combustion At STP

Jul 11, 2025 by ADMIN 64 views

Calculating the Volume of Carbon Dioxide Produced by Ethene Combustion at STP

In this comprehensive guide, we will delve into the process of calculating the volume of carbon dioxide (CO2) produced during the complete combustion of ethene (C2H4) at standard temperature and pressure (STP). This is a fundamental concept in chemistry, particularly in the realm of stoichiometry and gas laws. Stoichiometry deals with the quantitative relationships between reactants and products in chemical reactions, while gas laws govern the behavior of gases under different conditions. Understanding these principles is crucial for predicting the outcome of chemical reactions and performing various calculations in chemistry and related fields.

Our focus will be on a specific scenario: the complete combustion of 2.40 grams of ethene. We will systematically break down the steps involved in determining the volume of CO2 generated, including balancing the chemical equation, calculating the molar mass of reactants and products, using the mole concept to establish the relationship between ethene and CO2, and finally applying the ideal gas law to calculate the volume at STP. This detailed approach will not only provide a solution to the problem but also enhance your understanding of the underlying chemical principles.

By the end of this article, you will be equipped with the knowledge and skills to tackle similar stoichiometric calculations involving gas volumes. Whether you are a student learning chemistry or a professional in a related field, this guide will serve as a valuable resource for understanding and applying these concepts. So, let's embark on this journey of chemical calculations and unravel the intricacies of ethene combustion and CO2 production.

1. Balanced Chemical Equation: The Foundation of Stoichiometry

The very first step in any stoichiometric calculation is to write out the balanced chemical equation for the reaction. The balanced equation provides crucial information about the molar ratios of reactants and products, which is essential for determining the amount of product formed from a given amount of reactant. In our case, the reaction is the complete combustion of ethene (C2H4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The unbalanced equation is:

C2H4(g) + O2(g) → CO2(g) + H2O(l)

To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. We start by balancing the carbon atoms. There are 2 carbon atoms in ethene (C2H4), so we need 2 molecules of CO2 on the product side:

C2H4(g) + O2(g) → 2CO2(g) + H2O(l)

Next, we balance the hydrogen atoms. There are 4 hydrogen atoms in ethene, so we need 2 molecules of water (H2O) on the product side:

C2H4(g) + O2(g) → 2CO2(g) + 2H2O(l)

Finally, we balance the oxygen atoms. There are 4 oxygen atoms in 2 molecules of CO2 and 2 oxygen atoms in 2 molecules of H2O, giving a total of 6 oxygen atoms on the product side. Therefore, we need 3 molecules of oxygen (O2) on the reactant side to balance the equation:

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

This is the balanced chemical equation for the complete combustion of ethene. It tells us that 1 mole of ethene reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 2 moles of water. This molar ratio (1:3:2:2) is the key to our subsequent calculations. Understanding the balanced chemical equation is paramount for accurate stoichiometric calculations, as it lays the groundwork for determining the quantitative relationships between reactants and products. The coefficients in the balanced equation represent the number of moles of each substance involved in the reaction, which will be used to calculate the amount of CO2 produced from the given amount of ethene.

2. Molar Mass Calculations: Converting Grams to Moles

Having established the balanced chemical equation, the next crucial step is to calculate the molar masses of the substances involved, specifically ethene (C2H4) and carbon dioxide (CO2). The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in the chemical formula of the substance. This conversion from grams to moles is essential because stoichiometric calculations are based on the mole concept, which relates the amount of a substance to the number of particles it contains.

For ethene (C2H4), the molar mass is calculated as follows:

There are 2 carbon atoms, each with an atomic mass of 12.0 g/mol.
There are 4 hydrogen atoms, each with an atomic mass of 1.0 g/mol.

Therefore, the molar mass of C2H4 is (2 × 12.0 g/mol) + (4 × 1.0 g/mol) = 24.0 g/mol + 4.0 g/mol = 28.0 g/mol.

Similarly, for carbon dioxide (CO2), the molar mass is calculated as follows:

There is 1 carbon atom with an atomic mass of 12.0 g/mol.
There are 2 oxygen atoms, each with an atomic mass of 16.0 g/mol.

Therefore, the molar mass of CO2 is (1 × 12.0 g/mol) + (2 × 16.0 g/mol) = 12.0 g/mol + 32.0 g/mol = 44.0 g/mol.

Now that we have the molar masses of ethene and carbon dioxide, we can convert the given mass of ethene (2.40 g) into moles. To do this, we divide the mass of ethene by its molar mass:

Moles of C2H4 = Mass of C2H4 / Molar mass of C2H4 Moles of C2H4 = 2.40 g / 28.0 g/mol = 0.0857 mol

This calculation tells us that 2.40 grams of ethene is equivalent to 0.0857 moles. This value is crucial because it allows us to use the stoichiometric coefficients from the balanced chemical equation to determine the number of moles of CO2 produced. The accurate calculation of molar masses and the conversion of mass to moles are fundamental steps in stoichiometry, ensuring that we can quantitatively relate the amount of reactants and products involved in the chemical reaction.

3. Stoichiometric Calculations: Moles of Ethene to Moles of Carbon Dioxide

With the balanced chemical equation and the number of moles of ethene calculated, we can now use stoichiometry to determine the number of moles of carbon dioxide (CO2) produced. Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It allows us to predict the amount of product formed from a given amount of reactant, or vice versa, based on the mole ratios established in the balanced chemical equation.

The balanced chemical equation for the combustion of ethene is:

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)

From this equation, we can see that 1 mole of ethene (C2H4) produces 2 moles of carbon dioxide (CO2). This mole ratio (1:2) is the key to our calculation. It tells us that for every mole of ethene that reacts, 2 moles of carbon dioxide are formed. We have already calculated that 2.40 g of ethene is equivalent to 0.0857 moles of C2H4.

To find the number of moles of CO2 produced, we multiply the number of moles of C2H4 by the mole ratio:

Moles of CO2 = Moles of C2H4 × (Moles of CO2 / Moles of C2H4) Moles of CO2 = 0.0857 mol C2H4 × (2 mol CO2 / 1 mol C2H4) Moles of CO2 = 0.0857 mol × 2 Moles of CO2 = 0.1714 mol

Therefore, the complete combustion of 2.40 g of ethene (0.0857 moles) produces 0.1714 moles of carbon dioxide. This stoichiometric calculation is a direct application of the mole concept and the mole ratios derived from the balanced chemical equation. It demonstrates how we can use the quantitative information provided by the balanced equation to predict the amount of product formed in a chemical reaction. This step is crucial for determining the theoretical yield of CO2 and is a fundamental aspect of stoichiometric calculations in chemistry.

4. Ideal Gas Law: Calculating the Volume of Carbon Dioxide at STP

Now that we have determined the number of moles of carbon dioxide (CO2) produced (0.1714 moles), the final step is to calculate the volume of this gas at standard temperature and pressure (STP). STP is a reference point used for gas measurements, defined as 0°C (273.15 K) and 1 atmosphere (101.325 kPa) of pressure. To perform this calculation, we will use the ideal gas law, a fundamental equation in chemistry that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas.

The ideal gas law is expressed as:

PV = nRT

Where:

P is the pressure of the gas (in atmospheres, atm).
V is the volume of the gas (in liters, L).
n is the number of moles of the gas (in moles, mol).
R is the ideal gas constant (0.0821 L atm / (mol K)).
T is the temperature of the gas (in Kelvin, K).

At STP, the pressure P is 1 atm, and the temperature T is 273.15 K. We have already calculated the number of moles n of CO2 produced as 0.1714 moles. The ideal gas constant R is a known value. Our goal is to solve for the volume V.

Rearranging the ideal gas law equation to solve for V, we get:

V = nRT / P

Now, we can plug in the values:

V = (0.1714 mol) × (0.0821 L atm / (mol K)) × (273.15 K) / (1 atm) V = 3.83 L

Therefore, the volume of carbon dioxide produced by the complete combustion of 2.40 g of ethene at STP is 3.83 liters. This calculation demonstrates the application of the ideal gas law in determining the volume of a gas under specific conditions. The ideal gas law is a powerful tool for understanding and predicting the behavior of gases, and it is widely used in chemistry and related fields. The result, 3.83 liters, represents the volume that 0.1714 moles of CO2 would occupy under standard temperature and pressure, providing a quantitative answer to our initial problem.

In summary, we have successfully calculated the volume of carbon dioxide (CO2) produced by the complete combustion of 2.40 grams of ethene (C2H4) at standard temperature and pressure (STP). This calculation involved several key steps, each building upon the previous one to arrive at the final answer. We began by writing and balancing the chemical equation for the reaction, which is the foundation of any stoichiometric calculation. The balanced equation provided the crucial mole ratios between reactants and products, allowing us to relate the amount of ethene consumed to the amount of CO2 produced.

Next, we calculated the molar masses of ethene and carbon dioxide, which enabled us to convert the given mass of ethene into moles. This conversion is essential because stoichiometric calculations are based on the mole concept. We then used the mole ratio from the balanced equation to determine the number of moles of CO2 produced from the calculated number of moles of ethene. This step demonstrated the direct application of stoichiometry in predicting the amount of product formed in a chemical reaction.

Finally, we employed the ideal gas law to calculate the volume of CO2 at STP. The ideal gas law is a fundamental equation that relates the pressure, volume, number of moles, and temperature of an ideal gas. By plugging in the known values for pressure, temperature, number of moles, and the ideal gas constant, we were able to solve for the volume of CO2. The result, 3.83 liters, represents the volume that 0.1714 moles of CO2 would occupy under standard conditions.

This comprehensive calculation highlights the importance of understanding and applying various chemical principles, including balancing chemical equations, calculating molar masses, stoichiometry, and the ideal gas law. These concepts are fundamental to chemistry and are widely used in various scientific and industrial applications. The ability to perform such calculations is crucial for anyone studying chemistry or working in a related field. The step-by-step approach outlined in this article provides a clear and systematic method for tackling similar problems involving gas volumes and stoichiometric calculations.