Chapter 12, Problem 12.143P

Interpretation Introduction

Interpretation:

The mass of boron nitride that can be prepared from $1.00\text{ metric ton}$ of boric acid is to be calculated.

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass. Also, the amounts of substances in a balanced chemical reaction are stoichiometrically equivalent to each other.

A limiting reagent is the one that is completely consumed in a chemical reaction. The amount of product formed in any chemical reaction has to be in accordance with the limiting reagent of the reaction.

Amount(mol) of excess reactant is reactant left after the formation of maximum amount(mol) of products.

The formula to calculate moles is as follows:

  $\text{Amount}\left(\text{mol}\right)=\frac{\text{mass}}{\text{molar mass}}$

The ideal gas equation can be expressed as follows:

  $PV=nRT$        (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

Answer to Problem 12.143P

The mass of boron nitride that can be prepared from $1.00\text{ metric ton}$ of boric acid is $2.98\times {10}^5\text{g}$.

Explanation of Solution

The reation of step 1 is as follows:

  $\text{B}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{NH}}_3\left(g\right)\to \text{B}{\left({\text{NH}}_2\right)}_3\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)$        (2)

The reaction of step 2 is as follows:

  $\text{B}{\left({\text{NH}}_2\right)}_3\left(s\right)\to \text{BN}\left(s\right)+2{\text{NH}}_3\left(g\right)$        (3)

Add equation (1) and (2) to obtain the overall reaction.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}\text{B}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{NH}}_3\left(g\right)\to \text{B}{\left({\text{NH}}_2\right)}_3\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)\\ \text{B}{\left({\text{NH}}_2\right)}_3\left(s\right)\to \text{BN}\left(s\right)+2{\text{NH}}_3\left(g\right)\end{array}}\\ \text{B}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{NH}}_3\left(g\right)\to \text{BN}\left(s\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)\end{array}$

The fractional yield of the overall reaction is calculated as follows:

  $\text{Fractional yield}=\left(\frac{\text{yield of step1}}{100\%}\right)\left(\frac{\text{yield of step}\text{2}}{100\%}\right)$        (4)

Substitute $85.5\%$ for yield of step 1 and $86.8\%$ for yield of step 2 in the equation (4).

  $\begin{array}{c}\text{Fractional yield}=\left(\frac{85.5\%}{100\%}\right)\left(\frac{86.8\%}{100\%}\right)\\ =0.74214\end{array}$

The formula to calculate the moles of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ is as follows:

  $\text{Moles of B}{\left(\text{OH}\right)}_{\text{3}}=\left[\frac{\text{given mass}\text{of B}{\left(\text{OH}\right)}_{\text{3}}}{\text{molecular mass of B}{\left(\text{OH}\right)}_{\text{3}}}\right]$        (5)

Substitute $1\text{t}$ for the mass of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ and $61.83\text{ g/mol}$ for the molar mass of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ in the equation (5).

  $\begin{array}{c}\text{Moles of B}{\left(\text{OH}\right)}_{\text{3}}=\left(\frac{1\text{t}}{61.83\text{ g/mol}}\right)\left(\frac{1000\text{kg}}{1\text{t}}\right)\left(\frac{1000\text{g}}{\text{1}\text{kg}}\right)\\ =1.6173379\times {10}^4\text{mol}\end{array}$

Rearrange the equation (1) to calculate the number of moles of ${\text{NH}}_3$.

  $n_{{\text{NH}}_3}=\frac{PV}{RT}$        (6)

Substitute the value $3.07\times {10}^3\text{kPa}$ for $P$, $275\text{ K}$ for $T$, $12.5{\text{m}}^3$ for $V$ and $0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}$ for $R$ in the equation (6).

  $\begin{array}{c}{n}_{{\text{NH}}_{\text{3}}}=\frac{\left(3.07\times {10}^3\text{kPa}\right)\left(\frac{\text{1}\text{atm}}{101.325\text{kPa}}\right)\left(12.5{\text{m}}^3\right)\left(\frac{1\text{L}}{{10}^{-3}{\text{m}}^3}\right)}{\left(0.0821\text{ L}\cdot \text{atm/K}\cdot \text{mol}\right)\left(275\text{ K}\right)}\\ =1.6774720\times {10}^4\text{mol}\end{array}$

The molar ratio of reactants is 1:1. Moles of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ is less and therefore $\text{B}{\left(\text{OH}\right)}_{\text{3}}$ is the limiting reagent.

The formula to calculate the mass of $\text{BN}$ is as follows:

  $\text{Mass of}\text{BN}=\left[\begin{array}{l}\text{moles of}\text{B}{\left(\text{OH}\right)}_{\text{3}}\left(\frac{1\text{mol}\text{BN}}{1\text{mol}\text{B}{\left(\text{OH}\right)}_{\text{3}}}\right)\left(\text{molar mass of BN}\right)\\ \left(\text{fractional yield of overall reaction}\right)\end{array}\right]$        (7)

Substitute $1.6173379\times {10}^4\text{mol}$ for moles of $\text{B}{\left(\text{OH}\right)}_{\text{3}}$, $0.74214$ for fractional yield of overall reaction and $24.82\text{ g/mol}$ for molar mass of $\text{BN}$ in the equation (7).

  $\begin{array}{c}\text{Mass of}\text{BN}=\left[\begin{array}{l}\left(1.6173379\times {10}^4\text{mol}\right)\left(\frac{1\text{mol}\text{BN}}{1\text{mol}\text{B}{\left(\text{OH}\right)}_{\text{3}}}\right)\left(24.82\text{ g/mol}\right)\\ \left(0.74214\right)\end{array}\right]\\ =2.97912\times {10}^5\text{g}\\ \approx 2.98\times {10}^5\text{g}\text{.}\end{array}$

Conclusion

The mass of boron nitride that can be prepared from $1.00\text{ metric ton}$ of boric acid is $2.98\times {10}^5\text{g}$.