Question

Use the balanced equation for the combustion of ethane to complete the table. \begin{tabular}{|l|c|c|c|c|} \hline \multicolumn{5}{|c|}{\(2 \mathrm{C}_{2} \mathrm{H}_{4}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\)} \\ \hline Initially mixed & 10 molecules & 20 molecules & 0 molecules & 0 molecules \\ \hline How much reacts & & & & \\ \hline Composition of final mixture & & & & \\ \hline \end{tabular}

Short answer

6 molecules of C2H4 and 20 molecules of O2 will react to yield 12 molecules of CO2 and 18 molecules of H2O. 4 Molecules of C2H4 will be left unreacted.

Step-by-step solution

Understanding the given chemical equation

The given chemical equation is \(2C_2H_4(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g)\). In this combustion reaction, 2 molecules of Ethene (C2H4) and 7 molecules of Oxygen gas (O2) are reacting to form 4 molecules of Carbon dioxide (CO2) and 6 molecules of water vapor (H2O).

Calculate the amount that will react

In the table, we are given that 10 molecules of C2H4 and 20 molecules of O2 are initially mixed. According to the reaction, 2 molecules of C2H4 react with 7 molecules of O2. So, for 10 molecules of C2H4, we need 35 molecules of O2. But we only have 20 molecules, so not all the C2H4 can react. This means that O2 is our limiting reactant. The amount of O2 that will react is 20 molecules and since the ratio of C2H4 to O2 is 2:7 in the reactions, this means only \((2/7) * 20 = 5.71 \approx 6\) molecules of C2H4 will react.

Calculate the composition of final product

6 molecules of C2H4 reacts with the 20 molecules of O2. According to the reaction, 2 molecules of C2H4 and 7 molecules of O2 will yield 4 molecules of CO2 and 6 molecules of H2O. So, we get \((4/2) * 6 = 12\) molecules of CO2 and \((6/2) * 6 = 18\) molecules of H2O. Also from the initial mixture, 4 molecules of C2H4 are left unreacted.

Key concepts

Understanding Ethene

Ethene, also known as ethylene, is a simple hydrocarbon, represented by the chemical formula \(C_2H_4\). It is a colorless gas with a slightly sweet smell, often used in the production of polyethylene plastics. In chemical reactions such as combustion, ethene acts as a fuel. During combustion, ethene reacts with oxygen to produce carbon dioxide and water. This reaction is exothermic, meaning it releases energy in the form of heat. A well-known property of ethene is its ability to undergo polymerization, which is where small molecules (monomers) join together to form a large molecule (polymer).

Ethene is unsaturated, meaning it contains a double bond between the carbon atoms. This makes it especially reactive in addition reactions, where atoms are added across the double bond. Its combustive reaction is essential for understanding how hydrocarbons burn and release energy.

Decoding the Limiting Reactant Concept

In any chemical reaction, the limiting reactant is the substance that is entirely consumed first, thus stopping the reaction from continuing forward. Identifying the limiting reactant in a reaction is crucial because it determines the maximum amount of product that can be formed.

In our provided problem, we have a reaction between ethene \(C_2H_4\) and oxygen \(O_2\). According to the balanced equation, 2 molecules of ethene react with 7 molecules of oxygen to produce products. Let's look at a few critical points for determining the limiting reactant:

• Calculate the reactant amounts based on the stoichiometry of the reaction.
• Compare available quantities: Here, we have 10 molecules of ethene and 20 molecules of oxygen.
• Since 35 molecules of oxygen are needed for all of the ethene to react, and only 20 are available, oxygen is the limiting reactant.

This means not all ethene can react due to the insufficient amount of oxygen present.

Exploring Molecular Stoichiometry

Molecular stoichiometry involves using a balanced chemical equation to determine the relationships between the quantities of reactants and products. It is the quantitative study of the relationships between the amounts of substances consumed and produced. This specific exercise provides an excellent example of applying stoichiometry to a combustion reaction.

The key points of stoichiometry in this reaction include:

• Balanced Equation: Understand that each molecule or mole derived from the balanced equation \(2C_2H_4 + 7O_2 \rightarrow 4CO_2 + 6H_2O\) signifies a specific ratio of reactants and products.
• Calculate Reactant and Product Quantities: For example, from the given amounts, I used the ratio from the equation to determine how much \(C_2H_4\) and \(O_2\) would react.
• Product Formation: Calculate the resulting number of molecules for each product using the ratios from the balanced equation.

In this reaction because \(2C_2H_4\) produce \(4CO_2\) and \(6H_2O\), after adjusting for the limiting reactant, we conclude on the amounts of products formed.