Problem 43 - Chemistry Principles and Practice

Question

Because of the high reactivity of the boron hydrides with oxygen, they were considered as possible solid rocket fuels. Write the equation for the reaction of oxygen with \(\mathrm{B}_{10} \mathrm{H}_{14}\) (a solid at room temperature).

Step-by-Step Solution

Answer

The balanced reaction is \( \mathrm{B}_{10} \mathrm{H}_{14} + 11 \mathrm{O}_2 \rightarrow 5 \mathrm{B}_2\mathrm{O}_3 + 7 \mathrm{H}_2\mathrm{O} \).

Step 1: Identify the Reactants and Products

The problem states that we need to write the equation for the reaction of oxygen with \( \mathrm{B}_{10} \mathrm{H}_{14} \). Here, \( \mathrm{B}_{10} \mathrm{H}_{14} \) is one of the reactants, and oxygen \( \mathrm{O}_2 \) is the other. The products of the reaction will be the oxides of boron \( \mathrm{B}_2\mathrm{O}_3 \) and water \( \mathrm{H}_2\mathrm{O} \) due to the combustion of a boron hydride.

Step 2: Write the Unbalanced Equation

Based on the identified reactants and products, we can write the unbalanced chemical equation: \[ \mathrm{B}_{10} \mathrm{H}_{14} + \mathrm{O}_2 \rightarrow \mathrm{B}_2\mathrm{O}_3 + \mathrm{H}_2\mathrm{O} \] This equation shows the components before we balance each element.

Step 3: Balance the Boron Atoms

There are 10 boron atoms in \( \mathrm{B}_{10} \mathrm{H}_{14} \) as a reactant. Therefore, we need to have 10 boron atoms on the product side, requiring 5 \( \mathrm{B}_2\mathrm{O}_3 \) molecules. The equation becomes: \[ \mathrm{B}_{10} \mathrm{H}_{14} + \mathrm{O}_2 \rightarrow 5 \mathrm{B}_2\mathrm{O}_3 + \mathrm{H}_2\mathrm{O} \]

Step 4: Balance the Hydrogen Atoms

The compound \( \mathrm{B}_{10} \mathrm{H}_{14} \) has 14 hydrogen atoms. To balance these, we need 7 \( \mathrm{H}_2\mathrm{O} \, molecules \), as each water molecule contributes 2 hydrogen atoms. The equation now reads: \[ \mathrm{B}_{10} \mathrm{H}_{14} + \mathrm{O}_2 \rightarrow 5 \mathrm{B}_2\mathrm{O}_3 + 7 \mathrm{H}_2\mathrm{O} \]

Step 5: Balance the Oxygen Atoms

Now count the oxygen atoms on the product side. 5 \( \mathrm{B}_2\mathrm{O}_3 \) contains 15 oxygen atoms and 7 \( \mathrm{H}_2\mathrm{O} \) contains 7 oxygen atoms, totaling 22 oxygen atoms needed. Therefore, you need 11 \( \mathrm{O}_2 \) molecules on the reactant side: \[ \mathrm{B}_{10} \mathrm{H}_{14} + 11 \mathrm{O}_2 \rightarrow 5 \mathrm{B}_2\mathrm{O}_3 + 7 \mathrm{H}_2\mathrm{O} \]

Step 6: Verify the Balanced Equation

Ensure that the number of each type of atom is equal on both sides of the equation. We have 10 boron, 14 hydrogen, and 22 oxygen atoms on both sides, confirming the equation is balanced. The final balanced equation is: \[ \mathrm{B}_{10} \mathrm{H}_{14} + 11 \mathrm{O}_2 \rightarrow 5 \mathrm{B}_2\mathrm{O}_3 + 7 \mathrm{H}_2\mathrm{O} \]

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in chemistry. It ensures that the same number of atoms of each element is present on both sides of the equation. This represents the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
When balancing equations, follow these simple steps:

• Write down the unbalanced equation with all reactants and products.
• List all the elements involved in the reaction.
• Count the number of atoms for each element in the reactants and products.
• Use coefficients to balance the elements one at a time, starting with the most complex molecule.
• Never change the subscripts of the chemical formulas, as they represent the substance's identity.

In the problem about boron hydrides, balancing means making sure we have equal numbers of boron, hydrogen, and oxygen atoms on both sides of the equation. This often involves adjusting coefficients methodically and double-checking calculations to avoid errors.

Combustion Reactions

Combustion reactions are a common type of chemical reaction where a substance combines with oxygen to produce oxides and release energy. They are exothermic, meaning they release heat and often light. Common examples include the burning of fossil fuels to generate energy.
Typically, a combustion reaction involves:

• A fuel (which could be a hydrocarbon or other combustible material).
• Oxygen as the oxidizing agent.
• Products that are generally oxides, which release energy during the process.

In the example problem, Boron hydride, or ext{ ext{B}_{10} ext{H}_{14}}, serves as the fuel, reacting with oxygen to form boron trioxide ( ext{ ext{B}_2 ext{O}_3}) and water ( ext{ ext{H}_{2} ext{O}}). Identifying the products of this chemical reaction is crucial for writing the correct chemical equation.

Boron Hydrides

Boron hydrides are compounds composed solely of boron and hydrogen. They are known for their unique structures, where boron atoms can form different geometric arrangements, often resulting in stable clusters. These compounds are of great interest due to their high energy content and potential applications in fuel and rocket technology.
Important properties of boron hydrides include:

• High reactivity, especially with oxygen, making them viable candidates for energy-rich applications.
• Ability to form a variety of stable isomers.
• Characteristics that include being either solids or gases at room temperature, depending on the specific compound.

The specific boron hydride in our problem, ext{ ext{B}_{10} ext{H}_{14}}, is a solid compound noted for its potential use in solid rocket fuels due to its reactive nature and ability to undergo combustion, forming stable oxides and water.