Problem 29 - Chemistry: An Atoms-Focused Approach

Question

There are two ways to write the equation for the combustion of ethane: $$\begin{aligned}\mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2} \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\\2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{H}_{2} \mathrm{O}(g)+4 \mathrm{CO}_{2}(g)\end{aligned}$$ Do the different ways of writing the equation affect the calculation of how much \(\mathrm{CO}_{2}\) is produced from a known quantity of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\)

Step-by-Step Solution

Answer

Based on the analysis and calculations, we can conclude that the two different ways of writing the equation for the combustion of ethane do not affect the calculation of the amount of CO₂ produced from a known quantity of C₂H₆. The mole ratio of C₂H₆ to CO₂ is the same (1:2) for both equations, and the amount of CO₂ produced remains the same (2 moles) when using either equation to calculate it for an example of 1 mole of C₂H₆.

Step 1: Analyze both equations

We have two equations for the combustion of ethane: 1. \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2}\mathrm{O}_{2}(g) \rightarrow 3\mathrm{H}_{2} \mathrm{O}(g)+2\mathrm{CO}_{2}(g)\) 2. \(2\mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{H}_{2} \mathrm{O}(g)+4 \mathrm{CO}_{2}(g)\) Both of these equations represent the same chemical process, but the first one has a fractional coefficient for \(\mathrm{O}_2\), while the second maintains integer coefficients by doubling the whole equation.

Step 2: Determine the mole ratio of \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{CO}_{2}\)

In order to calculate the amount of \(\mathrm{CO}_{2}\) produced from a known amount of \(\mathrm{C}_{2} \mathrm{H}_{6}\), we need to determine the mole ratio of \(\mathrm{C}_{2} \mathrm{H}_{6}\) to \(\mathrm{CO}_{2}\) from both equations. 1. In the first equation, the mole ratio of \(\mathrm{C}_{2} \mathrm{H}_{6}\) to \(\mathrm{CO}_{2}\) is 1:2. 2. In the second equation, the mole ratio of \(\mathrm{C}_{2} \mathrm{H}_{6}\) to \(\mathrm{CO}_{2}\) is 2:4, which also reduces to 1:2. Since the mole ratio of \(\mathrm{C}_{2} \mathrm{H}_{6}\) to \(\mathrm{CO}_{2}\) is the same for both equations (1:2), we can conclude that there will be no difference in the amount of \(\mathrm{CO}_{2}\) produced when using either equation to calculate it.

Step 3: Compare the calculations for an example quantity of \(\mathrm{C}_{2} \mathrm{H}_{6}\)

Let's assume we have 1 mole of \(\mathrm{C}_{2} \mathrm{H}_{6}\). 1. Using the first equation, we get: \(1\,\mathrm{mole}\, \mathrm{C}_{2} \mathrm{H}_{6} \times \frac{2\,\mathrm{moles}\, \mathrm{CO}_{2}}{1\,\mathrm{mole}\, \mathrm{C}_{2} \mathrm{H}_{6}} = 2\,\mathrm{moles}\, \mathrm{CO}_{2}\) 2. Using the second equation, we get: \(1\,\mathrm{mole}\, \mathrm{C}_{2} \mathrm{H}_{6} \times \frac{4\,\mathrm{moles}\, \mathrm{CO}_{2}}{2\,\mathrm{moles}\, \mathrm{C}_{2} \mathrm{H}_{6}} = 2\,\mathrm{moles}\, \mathrm{CO}_{2}\) In both calculations, the amount of \(\mathrm{CO}_{2}\) produced is the same (2 moles). Therefore, the different ways of writing these equations do not affect the calculation of how much \(\mathrm{CO}_{2}\) is produced from a known quantity of \(\mathrm{C}_{2} \mathrm{H}_{6}\).

Key Concepts

Stoichiometry

Stoichiometry is like the foundation of understanding chemical reactions. It helps us measure and predict the amounts of substances involved in a chemical reaction. At its core, stoichiometry is about using **balanced chemical equations** to relate quantities of reactants and products.
By knowing the proportions of each substance, it becomes easy to calculate how much of each one is needed or produced. For example, it tells us how many moles of carbon dioxide are produced from moles of ethane in a combustion reaction.
Stoichiometry relies on this concept of the **mole**, which is a standard scientific unit for measuring large quantities of very small entities like atoms, molecules, or other particles.

• Remember, stoichiometry is like a recipe for reactions, showing the exact proportions in which reactants combine and products form.
• These proportions are found using the coefficients in chemical equations.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They help us visualize the transformation of reactants into products. Each element in a chemical equation has a unique chemical symbol, and the transformation is signified by an arrow pointing from reactants to products.
In the equation for combusting ethane, for example, we see ethane and oxygen reacting to produce water and carbon dioxide. Chemical equations have to be balanced to comply with **the law of conservation of mass**, which states that matter cannot be created or destroyed.
Balancing ensures that the amount of each element is the same on both sides of the equation, as seen in the two versions of the combustion reaction provided.
While one version of the equation uses a fraction, the other uses whole numbers by doubling all terms involved. This shows that multiple balanced forms can exist for the same reaction.

• Fractions in coefficients can sometimes be used to save space or simplify equations, as long as the equation remains balanced.
• Balanced equations not only ensure the correct quantitative relationships but also reflect the actual path of a chemical reaction.

Mole Ratio

A mole ratio is directly derived from a balanced chemical equation and is essential in stoichiometric calculations. This ratio tells us the proportion of moles of one substance that react or are produced with respect to another.
In the combustion of ethane, for both equations, the mole ratio between ethane (\( \mathrm{C}_2\mathrm{H}_6 \)) and carbon dioxide (\( \mathrm{CO}_2 \)) remains 1:2. This means for each mole of ethane combusted, two moles of carbon dioxide form.
The mole ratio acts like a conversion factor that bridges reactant and product quantities. No matter if equations use fractions or integers, the fundamental mole ratio stays constant.

• Mole ratio is key in calculating the amount of product formed from a given amount of reactant.
• It simplifies predicting outcomes of reactions and is crucial for scaling reactions to different magnitude.

This consistent ratio is the reason why both equation forms yield the same amount of \( \mathrm{CO}_2 \) in calculations.