Problem 10 - Chemistry and Chemical Reactivity

Question

Ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) burns in oxygen. (a) What are the products of the reaction? (b) Write the balanced equation for the reaction. (c) What mass of \(\mathrm{O}_{2}\), in grams, is required for complete combustion of 13.6 of ethane? (d) What is the total mass of products expected from the combustion of 13.6 g of ethane?

Step-by-Step Solution

Answer

(a) Products: CO2 and H2O. (b) 2C2H6 + 7O2 → 4CO2 + 6H2O. (c) 50.637 g O2. (d) 64.237 g products.

Step 1: Determine Products of Reaction

When ethane (\(\text{C}_2\text{H}_6\)) burns in the presence of oxygen (\(\text{O}_2\)), it undergoes complete combustion. In complete combustion, hydrocarbons burn in oxygen to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)) as products.

Step 2: Write and Balance the Chemical Equation

Start with the unbalanced chemical equation: \[\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\]Balance the equation by ensuring there are equal numbers of each type of atom on both sides.Balanced Equation: \[2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}\]

Step 3: Calculate Moles of Ethane

The molar mass of ethane (\(\text{C}_2\text{H}_6\)) is calculated as follows: \ Carbon: \(2 \times 12.01\ g/mol\) \ Hydrogen: \(6 \times 1.008\ g/mol\) \Total molar mass = \(30.07\ g/mol\). \To find the moles of ethane from 13.6 g: \\(\text{moles of } \text{C}_2\text{H}_6 = \frac{13.6\ g}{30.07\ g/mol} = 0.4521\ mol\).

Step 4: Determine Moles of Oxygen Required

From the balanced equation, 2 moles of ethane react with 7 moles of oxygen. Use the ratio \\(\frac{7\ mol\ \text{O}_2}{2\ mol\ \text{C}_2\text{H}_6}\) to find moles of \(\text{O}_2\): \\(\text{moles of } \text{O}_2 = 0.4521\ mol \times \frac{7\ mol\ \text{O}_2}{2\ mol\ \text{C}_2\text{H}_6} = 1.5824\ mol \text{O}_2\).

Step 5: Calculate Mass of Required Oxygen

The molar mass of oxygen (\(\text{O}_2\)) is .\(32.00\ g/mol\).\Calculate the mass of oxygen required: \[\text{mass of } \text{O}_2 = 1.5824\ mol \times 32.00\ g/mol = 50.637\ g\]

Step 6: Calculate Total Mass of Products

According to the law of conservation of mass, total mass of products equals total mass of reactants. Ethane's mass is \(13.6\ g\) and oxygen's mass is \(50.637\ g\).\Thus, the total mass of products is \(13.6\ g + 50.637\ g = 64.237\ g\).

Key Concepts

Stoichiometry

Stoichiometry is the heart of understanding chemical reactions, particularly combustion. It's about the quantitative relationships between reactants and products in a chemical equation. In simpler terms, stoichiometry helps us determine how much of each substance we need to start with to produce a certain amount of product. It hinges on using the balanced chemical equation to establish these ratios.

For example, in the combustion of ethane, the balanced equation tells us that 2 moles of ethane require 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water. This provides a clear recipe, if you will, allowing chemists to predict how much oxygen is necessary and how much carbon dioxide and water will form. By analyzing the mole ratio derived from the balanced equation, we achieve precise predictions of quantities involved in chemical reactions.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show how reactants transform into products. The balanced equation not only shows the reactants and products but also the correct proportions of each chemical species involved in the reaction.

Writing a balanced chemical equation is crucial. For instance, the unbalanced equation for ethane combustion is \(\text{C}_2\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\). To balance it, you must have equal numbers of each type of atom on both sides of the equation. This results in \(2\text{C}_2\text{H}_6 + 7\text{O}_2 \rightarrow 4\text{CO}_2 + 6\text{H}_2\text{O}\). This ensures that the law of conservation of mass is obeyed, and no atoms are lost or gained during the reaction process.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It is a crucial concept in stoichiometry because it allows us to convert between the mass of a substance and the amount in moles.

To find the molar mass of a compound like ethane (\(\text{C}_2\text{H}_6\)), we sum the atomic masses of all atoms in a molecule. Ethane contains two carbon and six hydrogen atoms. Carbon has a molar mass of 12.01 g/mol, and hydrogen is 1.008 g/mol. Thus, ethane's molar mass is \(2 \times 12.01\ g/mol + 6 \times 1.008\ g/mol = 30.07\ g/mol\).

Once the molar mass is known, it helps in converting a given mass of ethane into moles, facilitating further stoichiometric calculations, as demonstrated in the exercise.

Conservation of Mass

The conservation of mass is a fundamental principle in chemistry. It states that mass in an isolated system is neither created nor destroyed during a chemical reaction. This principle is vital, as it implies that the mass of the reactants before the reaction will equal the mass of the products after the reaction.

For our ethane combustion example, we start with 13.6 g of ethane and 50.637 g of oxygen, combining to form a total product mass of 64.237 g. This demonstrates that the total mass remains constant throughout the reaction process, in alignment with the conservation of mass principle.

Understanding this concept allows students to accurately determine the amounts of reactants needed and the expected mass of the products despite how a substance changes form during a chemical reaction.