Problem 14 - Chemistry for Allied Health

Question

Write and balance combustion reactions for the following compounds. a. methane \(\left(\mathrm{CH}_{4}\right)\) b. propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) c. octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) d. ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\) e. sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\)

Step-by-Step Solution

Answer

Balanced equations: Methane: \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\) Propane: \(\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}\) Octane: \(\mathrm{C}_8\mathrm{H}_{18} + 12.5\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 9\mathrm{H}_2\mathrm{O}\) Ethanol: \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}\) Sucrose: \(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11} + 12\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 11\mathrm{H}_2\mathrm{O}\)

Step 1: Identify the Combustion Reaction

In a combustion reaction, a hydrocarbon or carbohydrate reacts with oxygen (02010) to produce carbon dioxide (03020) and water (032020). This generic form can be written as: \(\text{hydrocarbon} + O_2 \rightarrow CO_2 + H_2O\).

Step 2: Write the Unbalanced Equation

Write the unbalanced chemical equations for each compound. For example, for methane, the equation is: \ \(\mathrm{CH}_{4} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\) and follow the similar pattern for other compounds.

Step 3: Balance Carbon Atoms

Balance the number of carbon atoms on both sides of the equation by equating the number of \(\mathrm{CO}_{2}\) molecules to the number of carbon atoms in the hydrocarbon. Example for methane: \ \(\mathrm{CH}_4 + \mathrm{O}_2 \rightarrow \mathbf{1} \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}\).

Step 4: Balance Hydrogen Atoms

Balance the hydrogen atoms by adjusting the number of \(\mathrm{H}_2\mathrm{O}\) molecules to twice the number of hydrogen atoms in the original compound. Example for methane: \ \(\mathrm{CH}_4 + \mathrm{O}_2 \rightarrow 1 \mathrm{CO}_2 + \mathbf{2} \mathrm{H}_2\mathrm{O}\).

Step 5: Balance Oxygen Atoms

Count the total oxygen atoms on the product side (\(2\) from each \(\mathrm{CO}_2\) and \(1\) from each \(\mathrm{H}_2\mathrm{O}\)) and balance by changing the number of \(\mathrm{O}_2\) molecules on the reactant side. Example for methane: \ \(\mathrm{CH}_4 + \mathbf{2} \mathrm{O}_2 \rightarrow 1 \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O}\).

Step 6: Verify and Write Final Balanced Equations

Verify that each atom is balanced and write the final balanced equations:1. Methane: \(\mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\) 2. Propane: \(\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}\) 3. Octane: \(\mathrm{C}_8\mathrm{H}_{18} + 12.5\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 9\mathrm{H}_2\mathrm{O}\) 4. Ethanol: \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}\) 5. Sucrose: \(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11} + 12\mathrm{O}_2 \rightarrow 12\mathrm{CO}_2 + 11\mathrm{H}_2\mathrm{O}\).

Key Concepts

Balancing Equations

Balancing equations is essential in understanding chemical reactions, as it ensures that the Law of Conservation of Mass is adhered to. This law states that matter cannot be created or destroyed in a chemical reaction. To balance an equation, you must ensure that the number of each type of atom on the reactant side equals the number on the product side.

• Firstly, write down the unbalanced equation, showing the reactants turning into the products.
• Next, count the number of each type of atom on both sides.
• Start by balancing the atoms that appear in the least amount first.
• Adjust coefficients – the numbers in front of the molecules – not the subscripts, within the formulas.
• Repeat the process until all elements are balanced.

Understanding balancing becomes easier with practice, and it's a crucial skill especially for complex reactions such as hydrocarbon combustions.

Hydrocarbon Combustion

Hydrocarbon combustion involves the burning of compounds containing hydrogen and carbon to produce energy. Hydrocarbons can be found in fuels like methane, propane, and octane. When these compounds react with oxygen, they not only release energy but also produce carbon dioxide and water.
The general form of a hydrocarbon combustion reaction is: \[\text{hydrocarbon} + O_2 \rightarrow CO_2 + H_2O\]

• In a complete combustion, there is sufficient oxygen leading to carbon dioxide and water as products.
• Incomplete combustion happens when there is not enough oxygen, leading to by-products like carbon monoxide or soot.
• The equations must be balanced to reflect the correct stoichiometry of the reaction.

Hydrocarbon combustion is key in many applications, especially automotive and industrial processes, highlighting the importance of understanding these reactions.

Chemical Reactions

Chemical reactions are the process by which substances interact to form new substances. This process involves breaking bonds in the reactants and forming new bonds in the products.
Key characteristics of chemical reactions include:

• Reactants are the starting substances in a reaction, and products are the resulting substances.
• The reaction involves energy changes, often in the form of heat, light, or electricity.
• Indicators like color change, gas production, and temperature change can signal a chemical reaction.

Chemical reactions are categorized into types such as synthesis, decomposition, single replacement, double replacement, and combustion reactions. Combustion reactions specifically describe burning, as in hydrocarbon combustion, and are exothermic, releasing energy as heat.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is vital to predict the quantities of substances consumed and produced in a reaction. Stoichiometry is based on balanced equations which provide the ratios needed.

• The balanced equation tells you the molar relationship between reactants and products.
• To perform stoichiometric calculations, use the mole concept and Avogadro’s number.
• It allows chemists to calculate yields of reactions and scale reactions up or down.

For example, in the combustion of methane: \[ \mathrm{CH}_4 + 2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O} \]
The equation tells us that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water. Understanding stoichiometry is fundamental in fields ranging from laboratory research to industrial production.