Problem 76 - Chemistry The Molecular Nature of Matter

Question

Write balanced chemical equations for the complete combustion (in the presence of excess oxygen) of the following: (a) \(\mathrm{C}_{12} \mathrm{H}_{26}\) (a component of kerosene), (b) \(\mathrm{C}_{18} \mathrm{H}_{36}\) (a component of diesel fuel), (c) \(\mathrm{C}_{7} \mathrm{H}_{8}\) (toluene, a raw material in the production of the explosive TNT).

Step-by-Step Solution

Answer

(a) \(\mathrm{C}_{12}\mathrm{H}_{26} + 37\mathrm{O}_{2} \rightarrow 24\mathrm{CO}_{2} + 26\mathrm{H}_{2}\mathrm{O}\), (b) \(\mathrm{C}_{18}\mathrm{H}_{36} + 53\mathrm{O}_{2} \rightarrow 36\mathrm{CO}_{2} + 36\mathrm{H}_{2}\mathrm{O}\), (c) \(\mathrm{C}_{7}\mathrm{H}_{8} + 15\mathrm{O}_{2} \rightarrow 14\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}\).

Step 1: Write the Unbalanced Equations

Start by writing the unbalanced chemical equations for the complete combustion of each compound:(a) \(\mathrm{C}_{12}\mathrm{H}_{26} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)(b) \(\mathrm{C}_{18}\mathrm{H}_{36} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)(c) \(\mathrm{C}_{7}\mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)

Step 2: Balance the Carbon Atoms

Balance the carbon atoms first. Place coefficients in front of \(\mathrm{CO}_{2}\) equal to the number of carbon atoms in each compound:(a) \(\mathrm{C}_{12}\mathrm{H}_{26} + \mathrm{O}_{2} \rightarrow 12\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)(b) \(\mathrm{C}_{18}\mathrm{H}_{36} + \mathrm{O}_{2} \rightarrow 18\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)(c) \(\mathrm{C}_{7}\mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow 7\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\)

Step 3: Balance the Hydrogen Atoms

Next, balance the hydrogen atoms. Place coefficients in front of \(\mathrm{H}_{2}\mathrm{O}\) equal to half the number of hydrogen atoms in each compound:(a) \(\mathrm{C}_{12}\mathrm{H}_{26} + \mathrm{O}_{2} \rightarrow 12\mathrm{CO}_{2} + 13\mathrm{H}_{2}\mathrm{O}\)(b) \(\mathrm{C}_{18}\mathrm{H}_{36} + \mathrm{O}_{2} \rightarrow 18\mathrm{CO}_{2} + 18\mathrm{H}_{2}\mathrm{O}\)(c) \(\mathrm{C}_{7}\mathrm{H}_{8} + \mathrm{O}_{2} \rightarrow 7\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\)

Step 4: Balance the Oxygen Atoms

Lastly, balance the oxygen atoms. Count the total number of oxygen atoms needed for the carbon dioxide and water on the right side, and adjust the coefficient of \(\mathrm{O}_{2}\) accordingly.(a) \(\mathrm{C}_{12}\mathrm{H}_{26} + \frac{37}{2}\mathrm{O}_{2} \rightarrow 12\mathrm{CO}_{2} + 13\mathrm{H}_{2}\mathrm{O}\)(b) \(\mathrm{C}_{18}\mathrm{H}_{36} + \frac{53}{2}\mathrm{O}_{2} \rightarrow 18\mathrm{CO}_{2} + 18\mathrm{H}_{2}\mathrm{O}\)(c) \(\mathrm{C}_{7}\mathrm{H}_{8} + \frac{15}{2}\mathrm{O}_{2} \rightarrow 7\mathrm{CO}_{2} + 4\mathrm{H}_{2}\mathrm{O}\)

Step 5: Simplify the Equations

Since coefficients should be whole numbers, multiply each equation by 2 to eliminate the fractions:(a) \(\mathrm{C}_{12}\mathrm{H}_{26} + 37\mathrm{O}_{2} \rightarrow 24\mathrm{CO}_{2} + 26\mathrm{H}_{2}\mathrm{O}\)(b) \(\mathrm{C}_{18}\mathrm{H}_{36} + 53\mathrm{O}_{2} \rightarrow 36\mathrm{CO}_{2} + 36\mathrm{H}_{2}\mathrm{O}\)(c) \(\mathrm{C}_{7}\mathrm{H}_{8} + 15\mathrm{O}_{2} \rightarrow 14\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}\)

Key Concepts

Combustion Reactions

Combustion reactions are exothermic chemical reactions where a substance (typically a fuel) combines with an oxidizing element, usually oxygen, producing energy in the form of heat and light. For example, when hydrocarbon fuels like gasoline or kerosene burn in air, the carbon (C) and hydrogen (H) in these compounds react with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O).

In the case of kerosene, a component of which is the hydrocarbon \(\mathrm{C}_{12}\mathrm{H}_{26}\), the combustion reaction can be represented by the following balanced chemical equation:\[\mathrm{C}_{12}\mathrm{H}_{26} + 37\mathrm{O}_{2} \rightarrow 24\mathrm{CO}_{2} + 26\mathrm{H}_{2}\mathrm{O}\].
Understanding combustion reactions is crucial, not just for solving chemistry problems, but also in real-world applications such as fuel burning in engines, environmental impact assessments, and fire safety.

Chemical Stoichiometry

Chemical stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is the process of using balanced chemical equations as the basis for calculating how much product will form from a given amount of reactants, or vice versa.

For example, using the combustion of \(\mathrm{C}_{12}\mathrm{H}_{26}\) as a reference, the stoichiometry tells us that for every 1 mole of kerosene burned, 37 moles of oxygen are required to yield 24 moles of carbon dioxide and 26 moles of water. These ratios are essential when calculating the amount of fuel needed for a particular process or the emission levels resulting from combustion.
Furthermore, stoichiometry is foundational in fields like engineering, environmental science, and pharmacology, where precise measurements and chemical equations guide critical decisions.

Chemical Equation Balancing

Balancing chemical equations is the technique of equalizing the number of each element on both sides of the equation. This ensures that the Law of Conservation of Mass is obeyed, meaning that matter is neither created nor destroyed during a chemical reaction.

Let's take the combustion of toluene, \(\mathrm{C}_{7}\mathrm{H}_{8}\), which is used in the production of TNT. The balanced equation is: \[\mathrm{C}_{7}\mathrm{H}_{8} + 15\mathrm{O}_{2} \rightarrow 14\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}\].
The process of balancing starts with the compound that has the most complex (or the largest number of different) atoms, typically carbon, followed by hydrogen and then oxygen. The key to mastering this skill is practice and understanding that the coefficients in front of chemicals represent the moles of substance. This fundamental skill helps when predicting the results of reactions and is crucial for lab work, industrial processes, and in understanding the reaction behavior.