Question

Balance each of the following equations that describe combustion reactions. a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(C_{6} \mathrm{H}_{14}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) c. \(C_{6} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

The balanced combustion reactions are as follows: a. \(2\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+7\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2C_{6} \mathrm{H}_{14}(l)+19\mathrm{O}_{2}(g) \rightarrow 12\mathrm{CO}_{2}(g)+14\mathrm{H}_{2} \mathrm{O}(g)\) c. \(C_{6} \mathrm{H}_{12}(l)+9\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\)

Step-by-step solution

Balance the Carbon atoms

Firstly, let's balance the number of carbon atoms. There are 2 carbon atoms on the left side and only 1 on right. Therefore, we need to put a coefficient of 2 in front of CO₂. This gives us: a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Hydrogen atoms

Next, let's balance the hydrogen atoms. There are 6 hydrogen atoms on the left side and only 2 on the right. So, we need to put a coefficient of 3 in front of H₂O. This gives us: a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Oxygen atoms

Finally, we will balance the oxygen atoms. There are now a total of 7 oxygen atoms on the right side (4 from CO₂ and 3 from H₂O) and only 2 on the left side. Therefore, we need to put a coefficient of 3.5 (7/2) in front of O₂. However, we don't want to have a fraction as a coefficient, so we multiply all coefficients by 2 to get whole numbers: a. \(2\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+7\mathrm{O}_{2}(g) \rightarrow 4\mathrm{CO}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\) #Step 2: Balance Reaction b# b. \(C_{6} \mathrm{H}_{14}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Carbon atoms

We first balance the carbon atoms by putting a coefficient of 6 in front of CO₂: b. \(C_{6} \mathrm{H}_{14}(l)+\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Hydrogen atoms

Next, we balance the hydrogen atoms by putting a coefficient of 7 in front of H₂O: b. \(C_{6} \mathrm{H}_{14}(l)+\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g)+7\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Oxygen atoms

Finally, we balance the oxygen atoms by putting a coefficient of 19/2 in front of O₂. To avoid having a fraction, we can multiply all coefficients by 2: b. \(2C_{6} \mathrm{H}_{14}(l)+19\mathrm{O}_{2}(g) \rightarrow 12\mathrm{CO}_{2}(g)+14\mathrm{H}_{2} \mathrm{O}(g)\) #Step 3: Balance Reaction c# c. \(C_{6} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Carbon atoms

We first balance the carbon atoms by putting a coefficient of 6 in front of CO₂: c. \(C_{6} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Hydrogen atoms

Next, we balance the hydrogen atoms by putting a coefficient of 6 in front of H₂O: c. \(C_{6} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\)

Balance the Oxygen atoms

Finally, we balance the oxygen atoms by putting a coefficient of 9 in front of O₂: c. \(C_{6} \mathrm{H}_{12}(l)+9\mathrm{O}_{2}(g) \rightarrow 6\mathrm{CO}_{2}(g)+6\mathrm{H}_{2} \mathrm{O}(g)\)

Key concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. A classic example includes the burning of hydrocarbons, which are compounds composed of hydrogen and carbon, to produce carbon dioxide and water.

In the provided exercise, we're looking at the combustion of different hydrocarbons: ethanol (C₂H₅OH), hexane (C₆H₁₄), and an unsaturated hydrocarbon, represented by C₆H₁₂. Each of these hydrocarbons reacts with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O). To ensure the chemical equation reflects the principle of conservation of mass, the number of atoms of each element must be equal on both sides of the equation. This means balancing the chemical equation by adjusting coefficients, which represent the number of molecules involved in the reaction.

Many combustion reactions are used for energy generation, as in engines and power plants, while others are vital for biological processes, such as cellular respiration.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction. The basic principle behind stoichiometry is the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

When balancing chemical equations, stoichiometry comes into play as we use it to determine the correct proportions of reactants and products. For example, to balance a combustion reaction, we need to ensure that the amount of reactants equals the amount of products in terms of their respective quantities, typically measured in moles. This practice involves calculating the stoichiometric coefficients, which are the numbers placed in front of compounds in an equation to indicate the number of molecules or moles of that compound required to balance the equation.

In the exercise example, balancing the hydrocarbons' combustion requires a careful calculation of the stoichiometric coefficients to maintain the same number of each type of atom on both sides of the equation.

Chemical Reactions

Chemical reactions are processes where reactant substances are transformed into different products. During these reactions, the bonds between atoms are broken and reformed. There are different types of chemical reactions, and among them, the combustion reaction is just one example.

For a chemical reaction to be accurately represented, a chemical equation that encapsulates the change should be balanced. This entails the same number of each type of atom on the reactant side as on the product side, obeying the conservation of mass. Balancing chemical equations is an essential skill for any chemist because it reflects what actually happens during the reaction.

In teaching chemical reactions, it is important to emphasize that a balanced equation provides critical information, such as the proportion in which substances react and the proportion in which products are formed. This knowledge is fundamental for practical applications such as synthesizing materials, determining reaction yields, or even calculating the energy changes involved in the reaction.