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 L of ammonia at STP, assuming diborane is in excess?

Short answer

(a) The balanced chemical equations for borazine synthesis are: 1. \( \mathrm{B}_{2} \mathrm{H}_{6} + 6 \mathrm{NH}_{3} \rightarrow 2 (\mathrm{BH})_{3}(\mathrm{NH})_{3} + 6 \mathrm{H}_{2} \) 2. \( 6 \mathrm{LiBH}_{4} + 6 \mathrm{NH}_{4} \mathrm{Cl} \rightarrow 6 (\mathrm{BH})_{3}(\mathrm{NH})_{3} + 6 \mathrm{LiCl} + 12 \mathrm{H}_{2} \) (b) The Lewis dot structure of borazine is a hexagonal ring with alternating boron (B) and nitrogen (N) atoms connected by single and double bonds. Each boron atom bonds to one hydrogen (H) atom, and each nitrogen atom has one lone pair. (c) From 2.00 L of ammonia at STP and with diborane in excess, 2.42 grams of borazine can be prepared.

Step-by-step solution

Write balanced chemical equations for the two synthetic methods of producing borazine.

For the first reaction, we have diborane reacting with ammonia to form borazine and hydrogen: \[\mathrm{B}_{2} \mathrm{H}_{6} + 6 \mathrm{NH}_{3} \rightarrow 2 (\mathrm{BH})_{3}(\mathrm{NH})_{3} + 6 \mathrm{H}_{2}\] For the second reaction, we have lithium borohydride reacting with ammonium chloride to form borazine, lithium chloride, and hydrogen: \[6 \mathrm{LiBH}_{4} + 6 \mathrm{NH}_{4} \mathrm{Cl} \rightarrow 6 (\mathrm{BH})_{3}(\mathrm{NH})_{3} + 6 \mathrm{LiCl} + 12 \mathrm{H}_{2}\]

Draw the Lewis dot structure of borazine.

Borazine is an analog of benzene, with alternating boron and nitrogen atoms connected by single and double bonds in a hexagonal ring. Each boron atom is bonded to one hydrogen atom, while each nitrogen atom has one lone pair. The Lewis dot structure of borazine can be represented as follows: 

Calculate the mass of borazine produced from 2.00 L of ammonia at STP.

First, we need to determine the number of moles of ammonia present in 2.00 L at STP. To do this, we will use the ideal gas law formula: \[\text{PV} = \text{nRT}\] Since the ammonia is at STP (Standard Temperature and Pressure), we have P = 1 atm and T = 273.15 K. The ideal gas constant R = 0.0821 L*atm/mol*K. We can rearrange the formula to solve for the number of moles (n): \[\text{n} = \frac{\text{PV}}{\text{RT}}\] Substitute the values: \[\text{n} = \frac{(1 \: \text{atm})(2.00 \: \text{L})}{(0.0821 \: \text{L}*\text{atm}/\text{mol}*\text{K})(273.15 \: \text{K})} \approx 0.0899 \: \text{mol}\] From the balanced chemical equation in Step 1, we see that 6 moles of ammonia react to produce 2 moles of borazine. Therefore, we can use the stoichiometry to determine the moles of borazine produced: \[\text{Moles of borazine} = \frac{2 \: \text{moles of borazine}}{6 \: \text{moles of NH}_3} \times 0.0899 \: \text{mol} \approx 0.0300 \: \text{mol}\] Now, we can convert moles of borazine to grams. The molar mass of borazine is (3×(1.01) + 3×(14.0)) g/mol = 80.5 g/mol \[\text{Mass of borazine} = 0.0300 \: \text{mol} \times \frac{80.5 \: \text{g}}{1 \: \text{mol}} = 2.42 \: \text{g}\] So, 2.42 grams of borazine can be prepared from 2.00 L of ammonia at STP, assuming diborane is in excess.

Key concepts

Understanding Chemical Equations

Chemical equations play a pivotal role in describing chemical reactions. They are representations of the reactants and products in a reaction, depicting the process of chemical change. The balancing of chemical equations ensures that the law of conservation of mass is respected, meaning the number of atoms for each element remains the same before and after the reaction.

When writing balanced chemical equations, like those for the synthesis of borazine, it involves indicating the correct stoichiometric coefficients that balance the number of atoms for each element on both sides of the equation. For instance, the reaction of diborane with ammonia produces borazine and hydrogen. This equation, \[\mathrm{B}_{2} \mathrm{H}_{6} + 6 \mathrm{NH}_{3} \rightarrow 2 (\mathrm{BH})_{3}(\mathrm{NH})_{3} + 6 \mathrm{H}_{2}\], shows that one molecule of diborane reacts with six molecules of ammonia to form two molecules of borazine and twelve atoms of hydrogen. The coefficients 1, 6, 2, and 6 are proportional to the number of moles involved in the actual chemical reaction.

Lewis Dot Structures

Lewis dot structures are diagrams that represent the valence electrons of atoms within a molecule, crucial for predicting the bonding behavior and molecular geometry. They form the foundation for understanding molecular structure and the distribution of electrons among the atoms in a molecule.

For borazine, \( (\mathrm{BH})_{3}(\mathrm{NH})_{3} \), its Lewis structure shows alternating boron and nitrogen atoms bonded together in a hexagonal ring, similar to the arrangement in benzene. Each boron atom is attached to one hydrogen atom, and each nitrogen atom has a lone pair of electrons. It is important to note that double bonds are evenly distributed in the ring, indicating delocalized electrons, which contribute to the stability and resonance of the molecule. The ability to draw such Lewis dot structures is critical for students to visualize the molecule's electron arrangement and anticipate its reactivity and interaction with other substances.

Stoichiometry Calculations

Stoichiometry calculations are the quantitative aspect of chemistry that involves computing the amounts of reactants and products in a chemical reaction. These calculations depend on the balanced chemical equation and the relationships known as mole ratios.

For example, the question of how many grams of borazine can be synthesized from 2.00 L of ammonia at STP involves several stoichiometric steps. First, we apply the ideal gas law to find the moles of ammonia. Then, using the balanced equation, we establish a mole ratio to relate ammonia to borazine. Following this, we determine the moles of borazine that can be formed from the given quantity of ammonia. Lastly, we use the molar mass of borazine to convert these moles to grams, yielding the mass of borazine producible under the specified conditions. This way, stoichiometry helps in predicting the amounts of substances involved and guides the practical aspects of chemical synthesis.