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?

We are given that borazine \((\mathrm{BH})_{3}(\mathrm{NH})_{3}\) can be prepared from the reaction of diborane \((B_2H_6)\) with ammonia \((NH_3)\), with hydrogen \((H_2)\) as another product. The balanced chemical equation for this reaction is: \[B_2H_6 + 6NH_3 \rightarrow 2(\mathrm{BH}_3\mathrm{NH}_3) + 6H_2\]

The second reaction involves lithium borohydride \((LiBH_4)\) and ammonium chloride \((NH_4Cl)\) to produce borazine, with lithium chloride \((LiCl)\) and hydrogen \((H_2)\) as the other products. The balanced chemical equation for this reaction is: \[3LiBH_4 + 3NH_4Cl \rightarrow (\mathrm{BH}_3\mathrm{NH}_3) + 3LiCl + 8H_2\]

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.

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.