Question

Complete and balance each of the following combustion reactions. a. \(\mathrm{C}_{6} \mathrm{H}_{6}(l)+\mathrm{O}_{2}(g) \rightarrow\) b. \(\mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \rightarrow\) c. \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \rightarrow\)

Short answer

The short answers to the given combustion reactions are: a. \(C_6H_6(l) + 7.5O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)\) b. \(C_5H_{12}(l) + 8O_2(g) \rightarrow 5CO_2(g) + 6H_2O(l)\) c. \(C_{2}H_{6}O(l) + 1.5O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)\)

Step-by-step solution

Identify the products

In a combustion reaction, the products are CO2 and H2O. So, our products for this reaction will be CO2 and H2O. C6H6(l) + O2(g) -> CO2(g) + H2O(l)

Balance the equation

To balance the equation, we first need to balance the C atoms, followed by the H atoms, and finally the O atoms. C6H6(l) + O2(g) -> 6CO2(g) + 3H2O(l) Now, we have 6C, 6H, and 15O on the product side. To balance the O atoms, we need 15/2 moles of O2 on the reactant side. C6H6(l) + 7.5O2(g) -> 6CO2(g) + 3H2O(l) #b. C5H12(l) + O2(g) -> ???#

Identify the products

As mentioned earlier, the products for this combustion reaction will also be CO2 and H2O. C5H12(l) + O2(g) -> CO2(g) + H2O(l)

Balance the equation

We balance the equation similarly to the previous reaction. C5H12(l) + O2(g) -> 5CO2(g) + 6H2O(l) Now, we have 5C, 12H, and 16O on the product side. To balance the O atoms, we need 8 moles of O2 on the reactant side. C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(l) #c. C2H6O(l) + O2(g) -> ???#

Identify the products

For this combustion reaction, the products remain CO2 and H2O. C2H6O(l) + O2(g) -> CO2(g) + H2O(l)

Balance the equation

Balance the equation similarly to the first two reactions. C2H6O(l) + O2(g) -> 2CO2(g) + 3H2O(l) Now, we have 2C, 6H, and 7O on the product side. To balance the O atoms, we need 3.5 moles of O2 on the reactant side. C2H6O(l) + 1.5O2(g) -> 2CO2(g) + 3H2O(l) The final balanced equations are: a. C6H6(l) + 7.5O2(g) -> 6CO2(g) + 3H2O(l) b. C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(l) c. C2H6O(l) + 1.5O2(g) -> 2CO2(g) + 3H2O(l)

Key concepts

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry, particularly for combustion reactions. When balancing an equation, the goal is to ensure that the number of each type of atom on the reactant side is equal to the number on the product side. This is crucial because it follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

Here’s how you can approach this task:

• Begin by counting the number of atoms for each element in both the reactants and products.
• Start balancing with an element that appears in only one reactant and one product. This is often carbon (C) in hydrocarbon reactions.
• Next, balance hydrogen (H) atoms, which typically appear in water (H₂O).
• Lastly, balance oxygen (O) atoms. Since oxygen is often present in multiple compounds on each side, it's easier to adjust its coefficients at the end."

Balancing equations may require fractional coefficients, especially for oxygen. It’s perfectly acceptable to use fractions temporarily and then multiply all coefficients by the smallest number needed to convert them to whole numbers.

Hydrocarbon Combustion

Hydrocarbon combustion refers to the chemical reaction where a hydrocarbon burns in the presence of oxygen (O₂), typically producing carbon dioxide (CO₂) and water (H₂O) as the main products. This reaction is exothermic, meaning it releases heat, which is one reason why hydrocarbons are widely used as fuels.

In a complete combustion reaction:

• The hydrocarbon (such as C₆H₆) reacts with oxygen.
• This results in carbon atoms forming carbon dioxide, while hydrogen atoms form water.
• The overall conversion follows the general pattern: Hydrocarbon + O₂ → CO₂ + H₂O.

To ensure complete combustion, there must be a sufficient supply of oxygen. If oxygen is limited, incomplete combustion may occur, producing carbon monoxide (CO) and other by-products. Understanding combustion reactions is vital for applications in energy production and environmental science.

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products involved in chemical reactions. It is an essential tool for predicting the outcomes of reactions and ensuring the right proportions are used.

In the context of combustion reactions:

• The coefficients in a balanced equation indicate the molar ratios of the reactants and products.
• For example, in the balanced equation for benzene combustion, \( ext{C}_6 ext{H}_6 + 7.5 ext{O}_2 → 6 ext{CO}_2 + 3 ext{H}_2 ext{O} \), the ratio of benzene to oxygen is 1:7.5.
• This means that for every mole of benzene burned, 7.5 moles of oxygen are required to produce 6 moles of CO₂ and 3 moles of water.

Practicing stoichiometry helps in understanding not just how much of each substance is used or formed but enables one to scale reactions, crucial for industrial applications and laboratory settings. Remember, it’s like following a recipe but for molecules!