Question

Borazine, $(\mathrm{BH}{)}_3(\mathrm{NH}{)}_3,$ is an analog of ${\mathrm{C}}_6{\mathrm{H}}_6,$ benzene. It can be prepared from the reaction of diborane with ammonia, with hydrogen as another product; or from lithium borohydride and ammonium chloride, with lithium chloride and hydrogen as the other products. (a) Write balanced chemical equations for the production of borazine using both synthetic methods. (b) Draw the Lewis dot structure of borazine. (c) How many grams of borazine can be prepared from $2.00\mathrm{L}$ of ammonia at STP, assuming diborane is in excess?

Short answer

(a) Balanced chemical equations: 1. $B_2{H}_6+6N{H}_3\to 2({\mathrm{BH}}_3{\mathrm{NH}}_3)+6H_2$ 2. $3LiB{H}_4+3N{H}_{4}Cl\to ({\mathrm{BH}}_3{\mathrm{NH}}_3)+3LiCl+8H_2$ (b) Lewis dot structure of borazine: .. B - H : | : H - N - H : | : .. B - H : | : H - N - H (c) Mass of borazine prepared from 2.00 L of ammonia at STP: 2.40 grams

Step-by-step solution

Part (a) Balanced Chemical Equation for First Synthetic Method

We are given that borazine $(\mathrm{BH}{)}_3(\mathrm{NH}{)}_3$ can be prepared from the reaction of diborane $(B_2{H}_6)$ with ammonia $(N{H}_3)$, with hydrogen $(H_2)$ as another product. The balanced chemical equation for this reaction is: $$B_2{H}_6+6N{H}_3\to 2({\mathrm{BH}}_3{\mathrm{NH}}_3)+6H_2$$

Part (a) Balanced Chemical Equation for Second Synthetic Method

The second reaction involves lithium borohydride $(LiB{H}_4)$ and ammonium chloride $(N{H}_{4}Cl)$ to produce borazine, with lithium chloride $(LiCl)$ and hydrogen $(H_2)$ as the other products. The balanced chemical equation for this reaction is: $$3LiB{H}_4+3N{H}_{4}Cl\to ({\mathrm{BH}}_3{\mathrm{NH}}_3)+3LiCl+8H_2$$

Part (b) Lewis Dot Structure of Borazine

To draw the Lewis dot structure of borazine, we need to consider the arrangement of Boron, Nitrogen, and Hydrogen atoms with their respective valence electrons. The Lewis dot structure of borazine is as follows: .. B - H : | : H - N - H : | : .. B - H : | : H - N - H In borazine, each Boron atom is bonded to a Hydrogen atom and two Nitrogen atoms, and each Nitrogen atom is bonded to a Hydrogen atom and two Boron atoms.

Part (c) Mass of Borazine Prepared from 2.00 L of Ammonia at STP

We are given that 2.00 L of ammonia reacts at STP with diborane to produce borazine. Since diborane is in excess, ammonia is the limiting reagent. At STP, 1 mole of any gas occupies 22.4 L. Hence, we can find the moles of ammonia as follows: Moles of ammonia = volume (L) / molar volume at STP (L/mol) = 2.00 L / 22.4 L/mol = 0.08929 mol Using the balanced chemical equation, we can deduce that 6 mol of ammonia reacts with diborane to produce 2 mol of borazine. By proportion: Moles of borazine produced = (0.08929 mol NH_3) × (2 mol borazine / 6 mol NH_3) = 0.02976 mol borazine Now, let's find the molar mass of borazine. Borazine's molecular formula is $(\mathrm{BH}{)}_3(\mathrm{NH}{)}_3$. Molar mass of borazine = (3 × [1 (mass of H)] + 3 × [10.81 (mass of B)] + 3 × [14.01(mass of N)]) g/mol = 80.47 g/mol Finally, we can find the mass of borazine produced by multiplying moles of borazine by its molar mass: Mass of borazine = moles of borazine × molar mass of borazine = 0.02976 mol × 80.47 g/mol = 2.40 g Thus, 2.40 grams of borazine can be prepared from 2.00 L of ammonia at STP, assuming diborane is in excess.

Key concepts

Chemical Equations

Understanding chemical equations is crucial for describing reactions in chemistry. They represent a chemical reaction with symbols and formulas. For example, borazine synthesis involves two different reactions:

1. **Diborane and Ammonia Reaction**: The reaction between diborane ($B_2{H}_6$) and ammonia ($N{H}_3$) is written as$$B_2{H}_6+6N{H}_3\to 2(B{H}_{3}N{H}_3)+6H_2$$This equation indicates that one molecule of diborane reacts with six molecules of ammonia to produce two molecules of borazine and six molecules of hydrogen.

2. **Lithium Borohydride and Ammonium Chloride Reaction**: The other method uses lithium borohydride ($LiB{H}_4$) and ammonium chloride ($N{H}_{4}Cl$):$$3LiB{H}_4+3N{H}_{4}Cl\to (B{H}_{3}N{H}_3)+3LiCl+8H_2$$This equation shows that three molecules of lithium borohydride and ammonium chloride each produce one molecule of borazine, three molecules of lithium chloride, and eight molecules of hydrogen.

Balancing chemical equations is about ensuring that the number of each type of atom is the same on both sides of the equation. This reflects the law of conservation of mass, stating that matter cannot be created or destroyed in a closed system.

Lewis Dot Structure

To understand the Lewis dot structure of borazine, it's important to know the basics. The Lewis dot structure represents the arrangement of electrons in a molecule. They help us visualize the bonding between atoms.

Borazine, or $(\mathrm{BH}{)}_3(\mathrm{NH}{)}_3$, consists of alternating boron and nitrogen atoms forming a six-membered ring, similar to benzene. Here's how to think about it:

• *Boron (B) atoms* each have three valence electrons.
• *Nitrogen (N) atoms* each have five valence electrons.
• *Hydrogen (H) atoms* have one valence electron each.

In borazine, each boron binds to one nitrogen and one hydrogen, while each nitrogen is bonded to one boron and one hydrogen. The appearance resembles benzene but with alternating boron and nitrogen atoms in place of carbon atoms.

The complete structure of borazine features three boron atoms alternated with three nitrogen atoms, each bonded to hydrogen. This configuration reveals its stable, cyclic nature and resembles the electron sharing observed in aromatic compounds like benzene.

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions, letting us calculate the amounts of reactants and products. For borazine synthesis, stoichiometry allows us to determine how much product can be obtained from a given amount of reactant.

Let's focus on the reaction between diborane and ammonia:$$B_2{H}_6+6N{H}_3\to 2(B{H}_{3}N{H}_3)+6H_2$$This equation tells us that 6 molecules of ammonia react with 1 molecule of diborane to yield 2 molecules of borazine.

• When starting with a known volume or mass of a reactant, like 2.00 L of $N{H}_3$ at STP (standard temperature and pressure), you can apply stoichiometry to find the resulting amount of borazine.
• First, calculate the moles of ammonia, knowing 1 mole of gas occupies 22.4 L at STP. This results in approximately 0.08929 moles of $N{H}_3$.

Using the balanced equation, we see that every 6 moles of $N{H}_3$ produces 2 moles of borazine, informing us that 0.08929 mol $N{H}_3$ gives roughly 0.02976 mol borazine.

Stoichiometry effectively links the chemical equation to the real world, translating the balancing of reactions into measurable quantities in lab settings.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, expressed in grams per mole. Knowing this value is key to converting between grams of a substance and moles.

For borazine, or $(\mathrm{BH}{)}_3(\mathrm{NH}{)}_3$, we calculate its molar mass by summing the atomic masses of all the atoms in one molecule:

• Boron (B) has an atomic mass of approximately 10.81 g/mol.
• Nitrogen (N) has an atomic mass of approximately 14.01 g/mol.
• Hydrogen (H) has an atomic mass of 1.00 g/mol.

Given borazine’s formula, the calculation is:

$$3(1.00)+3(10.81)+3(14.01)=80.47extg/mol$$

This calculated molar mass ($80.47extg/mol$) lets you determine how many grams of borazine correspond to a specific number of moles produced, such as in our stoichiometry calculation of 0.02976 mol.Mass of borazine is found by multiplying this molar mass by the moles used:

$$0.02976\times 80.47=2.40extg$$

Therefore, having these calculations helps accurately connect lab experiments with theoretical chemical formulas, confirming the interaction between math and chemistry.