Problem 86 - Chemistry Matter and Change

Question

Combustion Write a chemical equation for the combustion of each of the following substances. If a compound contains carbon and hydrogen, assume that carbon dioxide gas and liquid water are produced. $$\begin{array}{l}{\text { a. solid barium }} \\ {\text { b. solid boron }} \\\ {\text { c. liquid acetone }\left(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\right)} \\ {\text { d. liquid octane }\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)}\end{array}$$

Step-by-Step Solution

Answer

Combustion equations: a. 2Ba + O2 → 2BaO; b. 4B + 3O2 → 2B2O3; c. C3H6O + 4O2 → 3CO2 + 3H2O; d. 2C8H18 + 25O2 → 16CO2 + 18H2O.

Step 1: Understanding Combustion

Combustion is a chemical reaction where a substance combines with oxygen to produce heat and usually light. For organic compounds, this often results in carbon dioxide and water.

Step 2: Combustion of Solid Barium

Write the combustion equation for barium (Ba): When barium combusts with oxygen, it forms barium oxide. The equation is: \[\text{2Ba}_{(s)} + \text{O}_{2(g)} \rightarrow \text{2BaO}_{(s)}\].

Step 3: Combustion of Solid Boron

Write the combustion equation for boron (B): When boron combusts with oxygen, it forms boron trioxide. The equation is: \[\text{4B}_{(s)} + \text{3O}_{2(g)} \rightarrow \text{2B}_{2}\text{O}_{3(s)}\].

Step 4: Combustion of Liquid Acetone

Write the combustion equation for acetone (C\(_3\)H\(_6\)O): In combustion, it reacts with oxygen to produce carbon dioxide and water. The equation is: \[\text{C}_{3}\text{H}_{6}\text{O}_{(l)} + \text{4O}_{2(g)} \rightarrow \text{3CO}_{2(g)} + \text{3H}_{2}\text{O}_{(l)}\].

Step 5: Combustion of Liquid Octane

Write the combustion equation for octane (C\(_8\)H\(_{18}\)): In combustion, octane reacts with oxygen to produce carbon dioxide and water. The balanced equation is: \[\text{2C}_{8}\text{H}_{18(l)} + \text{25O}_{2(g)} \rightarrow \text{16CO}_{2(g)} + \text{18H}_{2}\text{O}_{(l)}\].

Key Concepts

Chemical Equations

Chemical equations symbolize chemical reactions. They show the reactants, the initial substances, and the products, the final substances after the reaction has taken place. In a chemical equation, you'll typically see chemical formulas that represent the substances involved.

Let's take a common chemical reaction like combustion. In the combustion of a substance, we start with the original substance, often a hydrocarbon like octane, along with oxygen. The chemical equation will provide information on how these two react to produce new products, such as carbon dioxide and water. For example, the combustion of octane is represented as:

• Reactants: \( ext{2C}_8 ext{H}_{18(l)} + 25 ext{O}_{2(g)} \)
• Products: \( ext{16CO}_{2(g)} + 18 ext{H}_{2} ext{O}_{(l)} \)

Chemical equations are not only about numbers and formulas, they tell a story of how substances transform.

Substance Combustion

Substance combustion describes a reaction where a material burns in oxygen, releasing energy in the form of heat and light. Combustion is an exothermic process, meaning it releases energy. This makes combustion useful in various applications, from heating homes to powering cars.

When we talk about the combustion of substances in chemistry, we're often dealing with hydrocarbons - compounds made of hydrogen and carbon. But combustion can also involve elements like barium or boron. In these cases, they generally react with oxygen to form oxides. For instance:

• Barium combustion yields barium oxide: \( ext{2Ba}_{(s)} + ext{O}_{2(g)} ightarrow ext{2BaO}_{(s)} \)
• Boron combustion produces boron trioxide: \( ext{4B}_{(s)} + ext{3O}_{2(g)} ightarrow ext{2B}_{2} ext{O}_{3(s)} \)

Balancing Equations

Balancing chemical equations ensures the law of conservation of mass holds true. This law states that matter cannot be created or destroyed in a chemical reaction, meaning the mass of reactants must equal the mass of products.

To balance a chemical equation, adjust the coefficients in front of the chemical formulas so that the number of atoms of each element is the same on both sides of the equation. This might require a bit of trial and error. Take the combustion of acetone:

• Original reactants and products: \( ext{C}_{3} ext{H}_{6} ext{O}_{(l)} + ext{O}_{2(g)} ightarrow ext{CO}_{2(g)} + ext{H}_{2} ext{O}_{(l)} \)
• Balanced equation: \( ext{C}_{3} ext{H}_{6} ext{O}_{(l)} + ext{4O}_{2(g)} ightarrow ext{3CO}_{2(g)} + ext{3H}_{2} ext{O}_{(l)} \)

The key? Carefully count each type of atom on both sides of the equation and adjust the coefficients as necessary. This keeps the equation balanced.

Reaction Products

Reaction products are the substances that result from a chemical reaction. In a combustion reaction, particularly with hydrocarbons, the products are typically carbon dioxide and water.

These reaction products are important for several reasons. Carbon dioxide is a notable greenhouse gas, implicated in global warming, while water is vital for all forms of life.

Let's consider the reaction products from the combustion of octane:

• Reactants: \( ext{2C}_8 ext{H}_{18(l)} + 25 ext{O}_{2(g)} \)
• Reaction products: \( ext{16CO}_{2(g)} + 18 ext{H}_{2} ext{O}_{(l)} \)

Understanding what products are formed from chemical reactions helps predict the effects and potential uses of these reactions. For example, while combustion products are routinely studied for their environmental impacts, they also inform fields like energy production and materials science.