Q12.138P - Chemistry: Molecular Nature Of Matter And Change

Question

An aerospace manufacturer is building a prototype experimental aircraft that cannot be detected by radar. Boron nitride is chosen for incorporation into the body parts, and the boric acid/ammonia method is used to prepare the ceramic material. Given 85.5% and 86.8% yields for the two reaction steps, how much boron nitride can be prepared from 1.00 metric ton of boric acid and $\mathbf{12}\mathbf{.}\mathbf{5}\mathbf{ }{\mathbf{m}}^{\mathbf{3}}$ of ammonia at 275 K and $\mathbf{3}\mathbf{.}\mathbf{07}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\text{ kPa}}$ ? Assume that ammonia does not behave ideally under these conditions and is recycled completely in the reaction process.

Step-by-Step Solution

Answer

From the chemical equation, one mole of boric acid reacts with one mole of ammonia to form one mole of Boron nitride. Here, the $1.62\times {10}^6$ moles of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ form the $1.62\times {10}^6$ moles of BN.

Definition

Intermolecular hydrogen bonding is the bonding between the two different molecules that approach each other. Hydrogen bonding is the kind of weak intermolecular force of attraction between the electronegative atom and electropositive atom which can be of the same molecule or a different molecule.

 

Viscosity can be defined as the force of attraction between the atoms of the layers. The viscosity decides the flow of the liquid. If the viscosity is high, the flow of liquid is low as it is viscous whereas if the viscosity is low, the flow of liquid is high.

The chemical reaction.

Boron Nitride can be prepared from the chemical reaction of the Boron hydroxide and ammonia.

$$\begin{gathered} \text{Step1:} \mathrm{B}(\mathrm{OH}{)}_3\left(\mathrm{s}\right)+3{\mathrm{NH}}_3\left(\mathrm{g}\right)\to \mathrm{B}(\mathrm{NH}2{)}_3\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \\ \text{Step2} \text{:} \mathrm{B}({\mathrm{NH}}_2{)}_3\left(\mathrm{s}\right)\stackrel{\mathrm{\Delta }}{\to }\mathrm{BN}\left(\mathrm{s}\right)+2{\mathrm{NH}}_3\left(\mathrm{s}\right) \\ \text{Overall Reaction :} \mathrm{B}(\mathrm{OH}{)}_3+{\mathrm{NH}}_3\left(\mathrm{g}\right)\to \mathrm{BN}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right) \end{gathered}$$

 

From the above reaction, the yields are 85.5% for step 1 and 86.8% for step 2.

The Fraction Yield of reaction.

$$\begin{gathered} \mathrm{Fraction} \mathrm{Yield}=(\mathrm{Yield} \mathrm{For} \mathrm{Step} 1)\times (\mathrm{Yield} \mathrm{For} \mathrm{Step} 2) \\ \text{FractionYield}=\left(\frac{85.5\%}{100\%}\right)\left(\frac{86.8\%}{100\%}\right) \\ \text{FractionYield}=0.74 \end{gathered}$$

The limiting agent.

$$\begin{gathered} 1.00 \mathrm{ton}=1.00\times {10}^3 \mathrm{kg} \\ 1.00 \mathrm{kg}=1.00\times {10}^3 \mathrm{g} \\ 1.00 \mathrm{ton}\times \frac{1.00\times {10}^3 \mathrm{kg}}{1.00 \mathrm{ton}}\times \frac{1.00\times {10}^3 \mathrm{g}}{1.00 \mathrm{kg}}=1.00\times {10}^6 \mathrm{g} \end{gathered}$$

Molar Mass of  $\mathrm{B}(\mathrm{OH}{)}_3=61.83 \mathrm{g}/\mathrm{mole}$

 

Number of Moles of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$:

$$\begin{gathered} \mathrm{Mole} \mathrm{of} \mathrm{B}\left(\mathrm{OH}\right)3=\frac{\mathrm{Amount} \mathrm{of} \mathrm{B}{\left(\mathrm{OH}\right)}_3}{\mathrm{Molar} \mathrm{Mass} \mathrm{of} \mathrm{B}{\left(\mathrm{OH}\right)}_3} \\ \text{Mole} \text{ of} {\text{B(OH)}}_3=\frac{1.00\times {10}^6\mathrm{g}}{61.83\mathrm{g}/\text{mole}} \\ \text{Mole} \text{ of} {\text{B(OH)}}_3=1.62\times {10}^6 \text{mole} \mathrm{B}{\left(\mathrm{OH}\right)}_3 \end{gathered}$$

 

As,

$\text{Overall Reaction :} \mathrm{B}(\mathrm{OH}{)}_3+{\mathrm{NH}}_3\left(\mathrm{g}\right)\to \mathrm{BN}\left(\mathrm{s}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the chemical equation, one mole of boric acid reacts with one mole of ammonia to form one mole of Boron nitride. Here, the $1.62\times {10}^6$ moles of  $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ form the $1.62\times {10}^6$ moles of BN.