Question

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

Short answer

(a) \(2\text{CH}_4\text{O} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 4\text{H}_2\text{O}\) (b) \(2\text{C}_3\text{H}_8\text{O} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 8\text{H}_2\text{O}\)

Step-by-step solution

Write the Unbalanced Equation

Start by writing the unbalanced equation for each combustion reaction.(a) \( \text{CH}_4\text{O}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \) (b) \( \text{C}_3\text{H}_8\text{O}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g) \)

Balance Carbon Atoms

Look at the equation and balance the carbon atoms by adjusting the coefficient of \( \text{CO}_2 \) on the product side.(a) 1 carbon in \( \text{CH}_4\text{O} \) and \( \text{CO}_2 \), so the coefficients remain 1.(b) 3 carbons in \( \text{C}_3\text{H}_8\text{O} \) necessitates 3 \( \text{CO}_2 \). The equation becomes:\[ \text{C}_3\text{H}_8\text{O}(l) + \text{O}_2(g) \rightarrow 3\text{CO}_2(g) + \text{H}_2\text{O}(g) \]

Balance Hydrogen Atoms

Now balance the hydrogen atoms by adjusting the coefficient of \( \text{H}_2\text{O} \) on the product side.(a) 4 hydrogens in \( \text{CH}_4\text{O} \) requires 2 \( \text{H}_2\text{O} \). So:\[ \text{CH}_4\text{O}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \](b) 8 hydrogens in \( \text{C}_3\text{H}_8\text{O} \) requires 4 \( \text{H}_2\text{O} \). Updating it results in:\[ \text{C}_3\text{H}_8\text{O}(l) + \text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \]

Balance Oxygen Atoms

Finally, balance the oxygen atoms. Count the total oxygen atoms needed on the product side and adjust the \( \text{O}_2 \) on the reactant side accordingly.(a) The products need 4 oxygen atoms (2 for \( \text{CO}_2 \) and 2 for 2x \( \text{H}_2\text{O} \)). Adjust \( \text{O}_2 \):Thus:\[ \text{CH}_4\text{O}(l) + \frac{3}{2} \text{O}_2(g) \rightarrow \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \]To remove the fraction, multiply through by 2:\[ 2\text{CH}_4\text{O}(l) + 3 \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \](b) The products need 10 oxygen atoms (6 from 3x \( \text{CO}_2 \) and 4 from 4x \( \text{H}_2\text{O} \)). Adjust \( \text{O}_2 \) as follows:\[ \text{C}_3\text{H}_8\text{O}(l) + rac{9}{2} \text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \]Multiply by 2:\[ 2\text{C}_3\text{H}_8\text{O}(l) + 9\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 8\text{H}_2\text{O}(g) \]

Conclusion: Balanced Reactions

The final balanced combustion reactions by inspection are as follows:For CH₄O:\[ 2\text{CH}_4\text{O}(l) + 3 \text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 4\text{H}_2\text{O}(g) \]For C₃H₈O:\[ 2\text{C}_3\text{H}_8\text{O}(l) + 9\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 8\text{H}_2\text{O}(g) \]

Key concepts

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation. This reflects the conservation of mass principle, indicating that matter is neither created nor destroyed in chemical reactions. To balance a chemical equation:

• Start by writing the unbalanced chemical equation, including all reactants and products.
• Balance elements one by one, starting with those that are in the least amount and progressing to more abundant ones, like hydrogen and oxygen.
• Usually, carbon atoms are balanced first in combustion reactions, followed by hydrogen and then oxygen.
• If fractions appear, multiply the entire equation by an appropriate factor to clear them, ensuring all coefficients are whole numbers.
• Double-check to confirm that all atoms balance on either side of the equation.

Balancing equations may initially seem challenging, but with practice, it becomes an intuitive and systematic process.

Chemical Stoichiometry

Chemical stoichiometry involves using balanced chemical equations to calculate the quantities of reactants and products involved in a chemical reaction. It serves as a guide to understanding how substances interact during chemical reactions. Key aspects to consider include:

• The coefficients in a balanced chemical equation, which represent the ratio of molecules or moles of each substance.
• Calculating moles and using molar ratios to determine the amount of reactants needed or products formed.
• Utilizing Avogadro's number to relate the amount of moles to particles and molecules.
• Performing conversions between mass and moles using the molar mass of each substance.

By employing stoichiometry, we gain insight into the quantities and relationships of substances within reactions, facilitating problem-solving in chemical analysis and industry applications.

Hydrocarbon Combustion

Hydrocarbon combustion is a type of chemical reaction where a hydrocarbon reacts with oxygen gas. It's an exothermic reaction, meaning it releases energy in the form of heat and light. Common products of hydrocarbon combustion include carbon dioxide (CO₂) and water (H₂O). Here's how it works:

• Hydrocarbons consist of hydrogen and carbon. Examples include methane (CH₄), propane (C₃H₈), and alcohols like methanol (CH₄O) and propanol (C₃H₈O).
• During combustion, hydrocarbons break down and bond with oxygen to form water and carbon dioxide. The general formula is:

  \[\text{Hydrocarbon} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\]

• The efficiency of combustion depends on the availability of oxygen. Complete combustion occurs when there is an ample supply, ensuring all carbon forms CO₂.
• In cases of insufficient oxygen, incomplete combustion can occur, producing carbon monoxide (CO) and other hydrocarbons, which are less desirable due to lower energy generation and pollution concerns.

Understanding hydrocarbon combustion is essential for energy production, environmental science, and various industrial applications.