Question

Balance each of the following combustion reactions by inspection: (a) \(\mathrm{CH}_{4}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

(a) \(\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2O(g)\) (b) \(\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2O(g)\)

Step-by-step solution

Balance Carbon Atoms (a)

For the reaction \( ext{CH}_4(g) + ext{O}_2(g) \rightarrow ext{CO}_2(g) + ext{H}_2O(g)\), start by balancing the carbon atoms. There is 1 carbon atom in \( ext{CH}_4\) and 1 in \( ext{CO}_2\). The carbon atoms are already balanced.

Balance Hydrogen Atoms (a)

Now balance the hydrogen atoms. \( ext{CH}_4\) has 4 hydrogen atoms, while \( ext{H}_2O\) has 2. Thus, add a coefficient of 2 in front of \( ext{H}_2O\) to balance the hydrogens: \( ext{CH}_4(g) + ext{O}_2(g) \rightarrow ext{CO}_2(g) + 2 ext{H}_2O(g)\).

Balance Oxygen Atoms (a)

Balance the oxygen atoms next. The left side has a coefficient of 1 for \( ext{O}_2\), which means 2 oxygen atoms. The right side has 2 from \( ext{CO}_2\) and 2*1 from \( ext{H}_2O\), summing to 4. Thus, we need a coefficient of 2 in front of \( ext{O}_2\): \( ext{CH}_4(g) + 2 ext{O}_2(g) \rightarrow ext{CO}_2(g) + 2 ext{H}_2O(g)\).

Balance Carbon Atoms (b)

For the reaction \( ext{C}_3 ext{H}_8(g) + ext{O}_2(g) \rightarrow ext{CO}_2(g) + ext{H}_2O(g)\), start by balancing carbon. There are 3 carbon atoms in \( ext{C}_3 ext{H}_8\) and 1 in each \( ext{CO}_2\). Put a coefficient of 3 in front of \( ext{CO}_2\): \( ext{C}_3 ext{H}_8(g) + ext{O}_2(g) \rightarrow 3 ext{CO}_2(g) + ext{H}_2O(g)\).

Balance Hydrogen Atoms (b)

Now balance the hydrogen atoms. \( ext{C}_3 ext{H}_8\) has 8 hydrogens, while \( ext{H}_2O\) has 2. Add a coefficient of 4 in front of \( ext{H}_2O\): \( ext{C}_3 ext{H}_8(g) + ext{O}_2(g) \rightarrow 3 ext{CO}_2(g) + 4 ext{H}_2O(g)\).

Balance Oxygen Atoms (b)

Balance the oxygen atoms. The right side has 3*2 (from \( ext{CO}_2\)) + 4*1 (from \( ext{H}_2O\)) = 6 + 4 = 10 oxygen atoms. Put a coefficient of 5 in front of \( ext{O}_2\): \( ext{C}_3 ext{H}_8(g) + 5 ext{O}_2(g) \rightarrow 3 ext{CO}_2(g) + 4 ext{H}_2O(g)\).

Key concepts

Combustion Reactions

Combustion reactions are processes where a substance, usually a hydrocarbon, reacts with oxygen to produce carbon dioxide, water, and energy in the form of heat or light. These reactions are fundamental to various fields, from energy production to everyday tasks like driving a car.
During the combustion of hydrocarbons, the carbon and hydrogen atoms in the fuel are oxidized by the oxygen in the air. This means that carbon is transformed into carbon dioxide \(\mathrm{CO}_2\) and hydrogen into water \(\mathrm{H}_2O\). A classic example of a combustion reaction is burning methane (\(\mathrm{CH}_4\)):

• Original: \(\mathrm{CH}_4(g) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + \mathrm{H}_2O(g)\)
• Balanced: \( \mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + 2\mathrm{H}_2O(g)\)

These reactions release a large amount of energy because breaking of bonds in oxygen molecules and formation of new bonds in carbon dioxide and water is exothermic. Ensuring the balance of atoms—particularly carbon, hydrogen, and oxygen—is key to correctly writing these chemical equations.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's essential for predicting how much of each substance you need or can expect from a reaction.
In the context of balancing chemical equations, stoichiometry ensures that the same number of each type of atom is present on both sides of the equation. This follows the Law of Conservation of Mass, which states that matter is neither created nor destroyed.

• For the reaction involving propane \(\mathrm{C}_3\mathrm{H}_8\), balance carbon atoms by placing a coefficient of 3 in front of \(\mathrm{CO}_2\), addressing the three carbon atoms in propane.
• Next, balance hydrogen by adding a coefficient of 4 in front of \(\mathrm{H}_2O\) to comply with the 8 hydrogen atoms in \(\mathrm{C}_3\mathrm{H}_8\).
• Finally, oxygen can be balanced, considering these adjustments, requiring a coefficient of 5 in front of \(\mathrm{O}_2\).

This precise quantitative approach allows chemists to work with complex reactions accurately, ensuring all parts of the equation fulfill the conservation of atoms.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, and this transformation is depicted using chemical equations. Each reaction has reactants that undergo chemical changes to form new products.
A chemical reaction is foundational to understanding how substances interact. Basic types include synthesis, decomposition, single replacement, and double replacement reactions. However, combustion reactions, as discussed previously, are specifically a type of reaction where oxygen is a reactant, and energy is typically released.
Balancing chemical equations is a fundamental skill. It ensures that equations adhere to the conservation laws and represent the actual physical processes occurring. Considerations in a typical chemical reaction:

• Identify and write down the correct formula for each reactant and product.
• Count the number of atoms for each element on both sides of the equation.
• Adjust coefficients to get the same number of each type of atom on both sides.

Understanding these basic principles provides a better grasp of chemical phenomena and how they are represented symbolically through equations.