Question

If \(2.0 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}, 2.0 \mathrm{~mol} \mathrm{C}_{4} \mathrm{H}_{10},\) and \(2.0 \mathrm{~mol}\) \(\mathrm{C}_{6} \mathrm{H}_{6}\) are completely combusted in oxygen, which one produces the largest number of moles of \(\mathrm{H}_{2} \mathrm{O}\) ? Which one produces the least? Explain.

Short answer

The compound that produces the largest number of moles of H₂O upon complete combustion is C₄H₁₀ (butane), with 10.0 moles of H₂O, while the compound that produces the least number of moles of H₂O is C₆H₆ (benzene), with 6.0 moles of H₂O.

Step-by-step solution

Write balanced chemical equations for the combustion reactions

For each compound, we can write a general equation for its combustion reaction: Compound + O₂ → CO₂ + H₂O Then, we balance the equations: 1. For the combustion of CH₃CH₂CH₂COOH (butanoic acid): \( C_4H_8O_2 + O_2 \rightarrow CO_2 + H_2O \) Balancing this equation, we get: \( C_4H_8O_2 + 6O_2 \rightarrow 4CO_2 + 4H_2O \) 2. For the combustion of C₄H₁₀ (butane): \( C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \) Balancing this equation, we get: \( 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \) 3. For the combustion of C₆H₆ (benzene): \( C_6H_6 + O_₂ \rightarrow CO₂ + H₂O \) Balancing this equation, we get: \( 2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O \)

Determine moles of H₂O produced by each compound

Now, for each combustion reaction equation, we calculate the number of moles of H₂O produced: 1. For the combustion of 2.0 moles of CH₃CH₂CH₂COOH: \( 2.0\,\text{mol}\,C_4H_8O_2 \times \frac{4\,\text{mol}\,H_2O}{1\,\text{mol}\,C_4H_8O_2} = 8.0\,\text{mol}\,H_2O \) 2. For the combustion of 2.0 moles of C₄H₁₀: \( 2.0\,\text{mol}\,C_4H_{10} \times \frac{10\,\text{mol}\,H_2O}{2\,\text{mol}\,C_4H_{10}} = 10.0\,\text{mol}\,H_2O \) 3. For the combustion of 2.0 moles of C₆H₆: \( 2.0\,\text{mol}\,C_6H_6 \times \frac{6\,\text{mol}\,H_2O}{2\,\text{mol}\,C_6H_6} = 6.0\,\text{mol}\,H_2O \)

Compare moles of H₂O to find the largest and smallest producers

The number of moles of H₂O produced in each case are: - For CH₃CH₂CH₂COOH: 8.0 moles of H₂O - For C₄H₁₀: 10.0 moles of H₂O - For C₆H₆: 6.0 moles of H₂O Therefore, the compound that produces the largest number of moles of H₂O is C₄H₁₀ (butane), and the compound that produces the least number of moles of H₂O is C₆H₆ (benzene).

Key concepts

Understanding Stoichiometry

Stoichiometry is the branch of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's like a recipe for chemical equations, ensuring that everything balances. Using stoichiometry, we determine how different elements and compounds will interact and what they will produce.

Consider a combustion reaction, where a compound reacts with oxygen to produce carbon dioxide and water. To calculate the amount of water produced, we rely on stoichiometry. This involves using the coefficients from a balanced chemical equation to relate the quantities of one substance to another.

Let's break it down:

• Identify the balanced chemical equation. For example, the complete combustion of butane requires this equation: \( 2C_4H_{10} + 13O_2 ightarrow 8CO_2 + 10H_2O \).
• Use mole ratios to connect the number of moles of reactants to the number of moles of products. The coefficients in the equation tell us that 2 moles of butane produce 10 moles of water.

With this understanding, you can calculate exactly how much of each product is formed from a given amount of reactant.

Deciphering Chemical Equations

Chemical equations are symbolic representations of chemical reactions where the substances involved are expressed with their chemical formulas. These equations are vital for visualizing the transformation of molecules in a reaction.

Here's what you need to know to decipher a chemical equation:

• The left side contains the reactants, and the right side lists the products.
• The arrow signifies the direction of the reaction.
• Coefficients before each formula show the molar quantities needed.

Every atom on the reactant side must appear equally on the product side, a principle known as "conservation of mass." This leads us to balance equations carefully.

For example, in butanoic acid's combustion, the unbalanced equation simply looks like: \( C_4H_8O_2 + O_2 ightarrow CO_2 + H_2O \).

Through balancing, it becomes: \( C_4H_8O_2 + 6O_2 ightarrow 4CO_2 + 4H_2O \). This balanced equation ensures that all atoms from reactants reappear intact in the products. Such equations allow chemists to accurately predict the outcomes of reactions, making chemical equations fundamental in chemistry.

Mastering the Mole Concept

The mole is a fundamental concept in chemistry, serving as a bridge between the atomic and macroscopic worlds. It's a unit of measurement that allows chemists to count particles, such as atoms, molecules, or ions, in a given substance.

One mole is defined as exactly \( 6.022 \times 10^{23} \) of these entities. This immense number, known as Avogadro's number, enables chemists to perform calculations that involve enormous quantities of atoms and molecules.

In terms of combustion, like our example with benzene \((C_6H_6)\), understanding moles helps determine product amounts. For instance, in the reaction \( 2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O \), moles clarify conversions. If you start with 2.0 moles of benzene, you can use its mole ratio with water (from the balanced equation) to calculate that 6.0 moles of water will result.

Grasping the mole concept is essential for conversions between mass, number of particles, and volume, making it a cornerstone of stoichiometry and chemical reactions.