Problem 88 - Chemistry The Central Science In Si Units

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.

Step-by-Step Solution

Answer

Butane (\( \mathrm{C}_4\mathrm{H}_{10} \)) produces the most \( \mathrm{H}_2\mathrm{O} \) and benzene (\( \mathrm{C}_6\mathrm{H}_6 \)) the least.

Step 1: Determine the combustion reactions

Write the balanced chemical equation for each substance when combusted in oxygen. For \( \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} \) (butyric acid):\[ \mathrm{C}_4\mathrm{H}_8\mathrm{O}_2 + 5\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O} \]For \( \mathrm{C}_4\mathrm{H}_{10} \) (butane):\[ 2\mathrm{C}_4\mathrm{H}_{10} + 13\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 10\mathrm{H}_2\mathrm{O} \]For \( \mathrm{C}_6\mathrm{H}_6 \) (benzene):\[ 2\mathrm{C}_6\mathrm{H}_6 + 15\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} \]

Step 2: Analyze moles of H2O produced for each reaction

For \( \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} \), the reaction produces 4 moles of \( \mathrm{H}_2\mathrm{O} \) per mole of \( \mathrm{C}_4\mathrm{H}_8\mathrm{O}_2 \). Thus, 2.0 moles of \( \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} \) will produce:\[ 2.0 \times 4 = 8.0 \text{ moles of } \mathrm{H}_2\mathrm{O} \]For \( \mathrm{C}_4\mathrm{H}_{10} \), 1 mole produces 5 moles of \( \mathrm{H}_2\mathrm{O} \), therefore, 2.0 moles of \( \mathrm{C}_4\mathrm{H}_{10} \) will produce:\[ 2.0 \times 5 = 10.0 \text{ moles of } \mathrm{H}_2\mathrm{O} \]For \( \mathrm{C}_6\mathrm{H}_6 \), 1 mole produces 3 moles of \( \mathrm{H}_2\mathrm{O} \), thus 2.0 moles of \( \mathrm{C}_6\mathrm{H}_6 \) will produce:\[ 2.0 \times 3 = 6.0 \text{ moles of } \mathrm{H}_2\mathrm{O} \]

Step 3: Compare the moles of H2O produced

Comparing the moles of \( \mathrm{H}_2\mathrm{O} \) produced:- \( \mathrm{C}_4\mathrm{H}_{10} \) produces 10.0 moles- \( \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{COOH} \) produces 8.0 moles- \( \mathrm{C}_6\mathrm{H}_6 \) produces 6.0 molesThus, \( \mathrm{C}_4\mathrm{H}_{10} \) produces the most moles of \( \mathrm{H}_2\mathrm{O} \), and \( \mathrm{C}_6\mathrm{H}_6 \) produces the least.

Key Concepts

Stoichiometry

Stoichiometry is the science of calculating the relationships between reactants and products in a chemical reaction. It allows us to predict the amounts of products that will be produced from given amounts of reactants. In our exercise, we are dealing with the combustion of three different compounds, and stoichiometry will help us determine which compound produces the most and least water.
- A balanced chemical equation is essential in stoichiometry as it provides the ratio, called the mole ratio, of reactants to products. This ratio is crucial in calculating how much of one substance is needed or produced from another.
- In our case, the balanced equations for the combustion of butyric acid, butane, and benzene give us the information we need about the moles of water produced per mole of each compound combusted.
Understanding stoichiometry is vital for accurately calculating these ratios and solving chemical equations, leading to correct predictions of product formation.

Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants transforming into products, with reactants on the left and products on the right. Each compound is represented by a chemical formula, and coefficients indicate the number of moles of each substance. Balancing these equations is crucial for an accurate representation of the reaction.
For our combustion reactions:

• The equation for butyric acid: \[\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2 + 5\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}\] shows that 1 mole of butyric acid produces 4 moles of water.
• The equation for butane: \[2\mathrm{C}_4\mathrm{H}_{10} + 13\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 10\mathrm{H}_2\mathrm{O}\] indicates 2 moles of butane produce 10 moles of water.
• The equation for benzene: \[2\mathrm{C}_6\mathrm{H}_6 + 15\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}\] shows that 2 moles of benzene yield 6 moles of water.

Balancing is fundamental in ensuring the conservation of mass and allowing us to use these equations for accurate calculations.

Mole Calculations

Mole calculations involve determining how many moles of a substance are involved in a reaction based on its stoichiometric relationships. The mole is a standard scientific unit for measuring the amount of substance, and it is based on Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities per mole.
In our exercise, we calculate how many moles of water are produced in each reaction:

• For butyric acid: 2 moles of acid produce \(2 \times 4 = 8\) moles of water.
• For butane: 2 moles result in \(2 \times 5 = 10\) moles of water.
• For benzene: 2 moles yield \(2 \times 3 = 6\) moles of water.

Mole calculations provide a method to relate quantities of chemicals in a reaction, allowing us to compare which compound results in the most or least amount of a particular product, such as water, in our case.