Chapter 5, Problem 121E

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane:

$2{\text{NH}}_3\left(g\right)+3{\text{O}}_2\left(g\right)+2{\text{CH}}_4\left(g\right)\to 2\text{HCN}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

If 5.00 × 10³ kg each of NH₃, O₂, and CH₄ are reacted, what mass of HCN and of H₂O will be produced, assuming 100% yield?

Interpretation Introduction

Interpretation: The amount of $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$ that can be produced from the stated reaction is to be calculated.

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

To determine: The mass of $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$ that can be produced from the stated reaction.

Answer to Problem 121E

The mass of $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$ formed by the stated chemical reaction is $\underset{\_}{2.81\times 10^{3}kg}$ and $\underset{\_}{5.62\times 10^{3}kg}$ respectively.

Explanation of Solution

To determine: The mass of $\text{HCN}$ that can be produced from the stated reaction.

Given

The stated chemical reaction is,

${\text{2NH}}_3\left(g\right)+3{\text{O}}_2\left(g\right)+2{\text{CH}}_4\left(g\right)\to 2\text{HCN}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$

The given mass of ${\text{NH}}_3$ is $5.00\times {10}^3\text{kg}\left(5.00\times {10}^6\text{g}\right)$.

The given mass of ${\text{O}}_2$ is $5.00\times {10}^3\text{kg}\left(5.00\times {10}^6\text{g}\right)$.

The given mass of ${\text{CH}}_4$ is $5.00\times {10}^3\text{kg}\left(5.00\times {10}^6\text{g}\right)$.

The yield of the reaction is $100\%$.

The molar mass of ${\text{NH}}_3$ $\begin{array}{l}=\text{N}+\text{3H}\\ =\left(14+\left(3\times 1\right)\right)\text{g}/\text{mol}\\ =17\text{g}/\text{mol}\end{array}$

The molar mass of ${\text{O}}_2$ $\begin{array}{l}=2\text{O}\\ =\left(2\times 16\right)\text{g}/\text{mol}\\ =32\text{g}/\text{mol}\end{array}$

The molar mass of ${\text{CH}}_4$ $\begin{array}{l}=\text{C}+4\text{H}\\ =\left(12+\left(4\times 1\right)\right)\text{g}/\text{mol}\\ =16\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{the}\text{substance}}$

Substitute the value of the given mass and the molar mass of ${\text{NH}}_3$, ${\text{O}}_2$ and ${\text{CH}}_4$ in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{NH}}_3=\frac{5\times {10}^6\text{g}}{17\text{g}/\text{mol}}\\ =\text{0}\text{.294}\times {\text{10}}^6\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{O}}_2=\frac{5\times {10}^6\text{g}}{32\text{g}/\text{mol}}\\ =\text{0}\text{.156}\times {\text{10}}^6\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{CH}}_4=\frac{5\times {10}^6\text{g}}{16\text{g}/\text{mol}}\\ =\text{0}\text{.312}\times {\text{10}}^6\text{mol}\end{array}$

According to the stated reaction,

$2\text{mol}$ of ${\text{NH}}_3$ react with $2\text{mol}$ of ${\text{CH}}_4$.

$1\text{mol}$ of ${\text{NH}}_3$ reacts with $1\text{mol}$ of ${\text{CH}}_4$.

$\text{0}\text{.294}\times {\text{10}}^6\text{mol}$ of ${\text{NH}}_3$ react with $\text{0}\text{.294}\times {\text{10}}^6\text{mol}$ of ${\text{CH}}_4$.

The number of moles of ${\text{CH}}_4$ given are $\text{0}\text{.312}\times {\text{10}}^6\text{mol}$. Hence, it is not the limiting reactant.

According to the stated reaction,

$2\text{mol}$ of ${\text{NH}}_3$ react with $3\text{mol}$ of ${\text{O}}_2$.

$1\text{mol}$ of ${\text{NH}}_3$ reacts with ${\text{O}}_2$ $=\left(\frac{3}{2}\right)\text{mol}$

$\text{0}\text{.294}\times {\text{10}}^6\text{mol}$ of ${\text{NH}}_3$ reacts with ${\text{O}}_2$ $\begin{array}{l}=\left(\frac{3\times \text{0}\text{.294}\times {\text{10}}^6}{2}\right)\text{mol}\\ =\text{0}\text{.441}\times {\text{10}}^6\text{mol}\end{array}$

The number of moles of ${\text{O}}_2$ given are $\text{0}\text{.156}\times {\text{10}}^6\text{mol}$. Hence, it is the limiting reactant.

According to the stated reaction,

$3\text{mol}$ of ${\text{O}}_2$ produce $2\text{mol}$ of $\text{HCN}$.

$1\text{mol}$ of ${\text{O}}_2$ produces $\text{HCN}$ $=\left(\frac{2}{3}\right)\text{mol}$

$\text{0}\text{.156}\times {\text{10}}^6\text{mol}$ of ${\text{O}}_2$ produces $\text{HCN}$ $\begin{array}{l}=\left(\frac{2\times \text{0}\text{.156}\times {\text{10}}^6}{3}\right)\text{mol}\\ =0.104\times {10}^6\text{mol}\end{array}$

The molar mass of $\text{HCN}$ $\begin{array}{l}=\text{H}+\text{C}+\text{N}\\ =\left(1+12+14\right)\text{g}/\text{mol}\\ =27\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

Substitute the value of the number of moles of $\text{HCN}$ and the molar mass of $\text{HCN}$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}\text{HCN}=\left(\text{0}\text{.104}\times {\text{10}}^6\text{mol}\right)\times \left(27\text{g}/\text{mol}\right)\\ =2.8\times {10}^6\text{g}\\ =\underset{\_}{2.8\times 10^{3}kg}\end{array}$

To determine: The mass of ${\text{H}}_{\text{2}}\text{O}$ that can be produced from the stated reaction.

Given

The stated chemical reaction is,

${\text{2NH}}_3\left(g\right)+3{\text{O}}_2\left(g\right)+2{\text{CH}}_4\left(g\right)\to 2\text{HCN}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$

According to the stated reaction,

$3\text{mol}$ of ${\text{O}}_2$ produce $6\text{mol}$ of ${\text{H}}_2\text{O}$.

$1\text{mol}$ of ${\text{O}}_2$ produces ${\text{H}}_2\text{O}$ $=\left(\frac{6}{3}\right)\text{mol}$

$\text{0}\text{.156}\times {\text{10}}^6\text{mol}$ of ${\text{O}}_2$ produces ${\text{H}}_2\text{O}$ $\begin{array}{l}=\left(\frac{6\times \text{0}\text{.156}\times {\text{10}}^6}{3}\right)\text{mol}\\ =0.312\times {10}^6\text{mol}\end{array}$

The molar mass of ${\text{H}}_2\text{O}$ $\begin{array}{l}=2\text{H}+\text{O}\\ =\left(\left(2\times 1\right)+16\right)\text{g}/\text{mol}\\ =18\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

Substitute the value of the number of moles of ${\text{H}}_2\text{O}$ and the molar mass of ${\text{H}}_2\text{O}$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}{\text{H}}_2\text{O}=\left(\text{0}\text{.312}\times {\text{10}}^6\text{mol}\right)\times \left(18\text{g}/\text{mol}\right)\\ =5.62\times {10}^6\text{g}\\ =\underset{\_}{5.62\times 10^{3}kg}\end{array}$

Conclusion

The mass of $\text{HCN}$ and ${\text{H}}_{\text{2}}\text{O}$ formed by the stated chemical reaction is $\underset{\_}{2.81\times 10^{3}kg}$ and $\underset{\_}{5.62\times 10^{3}kg}$ respectively.