Question

Boron hydrides were once evaluated for possible use as rocket fuels. Complete and balance the following equation for the combustion of diborane. $$\mathrm{B}_{2} \mathrm{H}_{6}+\mathrm{O}_{2} \longrightarrow \mathrm{B}(\mathrm{OH})_{3}$$

Short answer

The balanced equation for the combustion of diborane (\(\mathrm{B}_{2}\mathrm{H}_{6}\)) with oxygen (\(\mathrm{O}_{2}\)) to produce boron hydroxide (\(\mathrm{B}(\mathrm{OH})_{3}\)) is: \[ \mathrm{B}_{2}\mathrm{H}_{6} + 3\mathrm{O}_{2} \longrightarrow 2\mathrm{B}(\mathrm{OH})_{3} \]

Step-by-step solution

Identify All Chemical Species

In this combustion reaction, we have: - Diborane (B2H6) as the reactant - Oxygen (O2) as the reactant - Boron hydroxide (BOH3) as the product

Write the Unbalanced Equation

Now, let's write the unbalanced chemical equation: $$ \mathrm{B}_{2}\mathrm{H}_{6} + \mathrm{O}_{2} \longrightarrow \mathrm{B}(\mathrm{OH})_{3} $$

Balance the Equation for Boron Atoms

Since there are 2 boron atoms in B2H6, we need 2 boron hydroxide (BOH3) units on the product side to balance the boron atoms: $$ \mathrm{B}_{2}\mathrm{H}_{6} + \mathrm{O}_{2} \longrightarrow \underline{2}\mathrm{B}(\mathrm{OH})_{3} $$

Balance the Equation for Hydrogen Atoms

We have 6 hydrogen atoms in B2H6, and there are 3 hydrogen atoms in each BOH3 molecule. Since we have 2 BOH3 molecules on the product side, the hydrogen atoms are already balanced: $$ \mathrm{B}_{2}\mathrm{H}_{6} + \mathrm{O}_{2} \longrightarrow 2\mathrm{B}(\mathrm{OH})_{3} $$

Balance the Equation for Oxygen Atoms

Finally, we need to balance the oxygen atoms. We have 3 oxygen atoms in each of the 2 BOH3 molecules on the product side, making a total of 6 oxygen atoms. So, we need 3 O2 molecules on the reactant side to balance the oxygen atoms: $$ \mathrm{B}_{2}\mathrm{H}_{6} + \underline{3}\mathrm{O}_{2} \longrightarrow 2\mathrm{B}(\mathrm{OH})_{3} $$

Check the Balanced Equation

Now our balanced equation is: $$ \mathrm{B}_{2}\mathrm{H}_{6} + 3\mathrm{O}_{2} \longrightarrow 2\mathrm{B}(\mathrm{OH})_{3} $$ Let's check the number of atoms for each element: - Boron: 2 atoms on both sides (B2H6 and 2*BOH3) - Hydrogen: 6 atoms on both sides (B2H6 and 2*(3*H)) - Oxygen: 6 atoms on both sides (3*O2 and 2*(3*O)) The chemical equation is now balanced.

Key concepts

Combustion Reaction

A combustion reaction is a type of chemical reaction where a substance combines with oxygen gas, releasing energy in the form of light and heat. Many substances, especially those containing carbon and hydrogen, undergo combustion. This process is common in everyday life, such as when fuels burn in a car engine or when a candle is lit. In the combustion reaction of diborane, both boron and hydrogen are oxidized in the presence of oxygen, producing boron hydroxide.

• Reactants typically include a fuel (such as diborane, which is a boron hydride) and oxygen.
• The products often include oxides like carbon dioxide, water, or in this case, boron hydroxide.

Combustion can be complete or incomplete. Complete combustion occurs when there is sufficient oxygen to allow the fuel to react completely to form products without any leftover reactants, mainly producing CO2 and H2O. Incomplete combustion may produce carbon monoxide or soot.

Diborane

Diborane (\( ext{B}_2 ext{H}_6 \)) is a covalent compound made of boron and hydrogen atoms. It’s interesting for its utilization in chemistry and potential as a high-energy rocket fuel.
This compound is colorless, highly flammable, and generates a lot of energy when it combusts.

• It consists of two boron atoms bonded to six hydrogen atoms.
• Despite its potential as a fuel, diborane's high reactivity makes it dangerous to handle.

In historical contexts, diborane was considered as rocket fuel due to the extensive energy released during its combustion. The molecules compact structure contributes to its energy density, but safety concerns have limited its practical use.

Stoichiometry

Stoichiometry is a branch of chemistry involving the calculation of reactants and products in chemical reactions. It's like a recipe that ensures the correct amounts of ingredients are mixed together.
In stoichiometry, understanding the molar ratio of a balanced chemical equation is essential, as it allows you to calculate how much of each reactant you need and what you will get as products.

• Chemists use stoichiometry to scale reactions up or down, based on the amount of starting substances.
• It is crucial for lab work, industry, and any scenario requiring precise chemical formulations.

For a balanced reaction, stoichiometry ensures that mass is conserved, meaning the mass and number of atoms of each element should remain equal on both sides of the reaction equation.

Balance Equation Steps

Balancing chemical equations is a necessary skill for chemists, as it ensures that the mass and energy in a chemical reaction are conserved. It might seem challenging at first, but by following systematic steps, it becomes much simpler.
Here is a streamlined approach to balancing equations:

• **Identify All Reactants and Products**: Determine the substances involved in the reaction's starting and ending states.
• **Write the Unbalanced Equation**: Note the chemical formula of the reactants and products.
• **Balance One Element at a Time**: Begin with an element that appears in only one reactant and one product, then proceed to balance other elements.
• **Adjust Coefficients, Not Subscripts**: Use coefficients in front of formulas to balance the equation, never alter the chemical formula itself.
• **Check Your Work**: Count the atoms of each element on both sides to ensure they are equal, confirming conservation of mass.

By iteratively adjusting coefficients and verifying the balance of atoms, one ensures the reaction abides by the law of conservation of mass.