Problem 75 - Chemistry The Molecular Nature of Matter

Question

Write balanced chemical equations for the complete compustion (in the presence of excess oxygen) of the following: (a) \(\mathrm{C}_{6} \mathrm{H}_{6}\) (benzene, an important industrial chemical and solvent), (b) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (butane, a fuel used in cigarette ighters), (c) \(\mathrm{C}_{21} \mathrm{H}_{44}\) (a component of paraffin wax used in candles).

Step-by-Step Solution

Answer

The balanced equations are: (a) \( 2\text{C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2O \), (b) \( 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2O \), (c) \( \text{C}_{21}\text{H}_{44} + 32\text{O}_2 \rightarrow 21\text{CO}_2 + 22\text{H}_2O \).

Step 1: Establish the General Form of a Combustion Reaction

In a combustion reaction, a hydrocarbon reacts with oxygen (\text{O}_2) to form carbon dioxide (CO_2) and water (H_2O). The general form of a complete combustion reaction is: \[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2O \]

Step 2: Balance the Combustion of Benzene (\text{C}_6\text{H}_6)

Write the unbalanced equation for benzene: \[ \text{C}_6\text{H}_6 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2O \] To balance the equation, we start by equating the number of carbon and hydrogen atoms on both sides: \[ 6\text{CO}_2 + 3\text{H}_2O \] Then we balance the oxygen atoms. There are a total of 15 oxygen atoms on the right side, requiring 7.5 \text{O}_2 molecules: \[ \text{C}_6\text{H}_6 + 7.5\text{O}_2 \rightarrow 6\text{CO}_2 + 3\text{H}_2O \] Since we cannot have fractional molecules, we multiply the entire equation by 2 to clear the fraction: \[ 2\text{C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2O \]

Step 3: Balance the Combustion of Butane (\text{C}_4\text{H}_{10})

Write the unbalanced equation for butane: \[ \text{C}_4\text{H}_{10} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2O \] To balance the equation, we match the number of carbon and hydrogen atoms on both sides first: \[ 4\text{CO}_2 + 5\text{H}_2O \] Now balance the oxygen atoms. There are 13 oxygen atoms on the right side, requiring 6.5 \text{O}_2 molecules: \[ \text{C}_4\text{H}_{10} + 6.5\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2O \] To clear the fraction, multiply the entire equation by 2: \[ 2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2O \]

Step 4: Balance the Combustion of Paraffin Component (\text{C}_{21}\text{H}_{44})

Write out the unbalanced equation: \[ \text{C}_{21}\text{H}_{44} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2O \] Balance carbon and hydrogen first: \[ 21\text{CO}_2 + 22\text{H}_2O \] Now balance the oxygen atoms. There are 64 oxygen atoms on the right side, so we need 32 \text{O}_2 molecules: \[ \text{C}_{21}\text{H}_{44} + 32\text{O}_2 \rightarrow 21\text{CO}_2 + 22\text{H}_2O \]

Key Concepts

Complete Combustion Reaction

The essence of a complete combustion reaction lies in the full utilization of a fuel, which is commonly a hydrocarbon, reacting with oxygen to yield carbon dioxide and water as products. This type of reaction occurs when there is an ample supply of oxygen (O₂), allowing the fuel to burn entirely without leaving unreacted fuel or producing byproducts such as carbon monoxide (CO) and soot.

A complete combustion reaction is typically represented by the general equation:
\[\text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2O\]
In the balanced chemical equation, every atom on the reactant side must be accounted for on the product side, upholding the law of conservation of mass. Thus, the balance of atoms determines the stoichiometric coefficients that precede each molecule in the equation, ensuring that the numbers of each type of atom are equal on both sides of the reaction.

Hydrocarbon Combustion

Hydrocarbon combustion is a chemical process involving a hydrocarbon—like benzene (C₆H₆), butane (C₄H₁₀), or paraffin components (C₂₁H₄₄)—reacting with oxygen to produce energy in the form of heat and light. This exothermic reaction powers our vehicles, heats buildings, and is used in many industrial processes.

Each hydrocarbon has a unique formula, which affects the balance of reactants and products in the combustion reaction. For instance:
\[2\text{C}_6\text{H}_6 + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2O\]
This balanced equation represents the combustion of benzene, where the coefficients indicate the precise amount of each chemical needed to satisfy the reaction's requirements.

Stoichiometry

Stoichiometry is the mathematics behind chemistry and is used for quantifying the relationships between reactants and products in chemical reactions. When balancing hydrocarbon combustion equations, it's pivotal to account for each atom of the elements involved, ensuring that the total mass of reactants equals the mass of products, as demanded by the law of conservation of mass.

Using stoichiometry, we can solve for quantities like the number of moles, the mass of the reactants needed, or the volume of gas produced. For example, the balanced equation for butane's combustion:
\[2\text{C}_4\text{H}_{10} + 13\text{O}_2 \rightarrow 8\text{CO}_2 + 10\text{H}_2O\]
illustrates how 2 moles of butane reacts with 13 moles of oxygen, producing 8 moles of carbon dioxide and 10 moles of water. These coefficients are vital for extrapolating the quantities needed in practical applications, such as determining the fuel requirements for thermal energy production.