Chapter 10, Problem 7E

The first step in the Ostwald process for manufacturing nitric acid is the reaction between ammonia and oxygen described by the equation ${\text{4NH}}_{\text{3}}+{\text{5O}}_{\text{2}}\to \text{4NO}+{\text{6H}}_{\text{2}}\text{O}$. Use this equation to answer all parts of this question.

a) How many moles of ammonia can be oxidized by $\text{268}\text{g}$ of oxygen?

b) If the reaction consumes $31.7$ moles of ammonia, how many grams of water will be produced?

c) How many grams of ammonia are required to produce $\text{404}\text{g}$ of nitrogen monoxide?

d) If $\text{6}\text{.41}\text{g}$ of water results from the reaction, what will be the yield of nitrogen monoxide (in grams)?

Interpretation Introduction

(a)

Interpretation:

The moles of ammonia that can be oxidized by $\text{268}\text{g}$ of oxygen are to be predicted.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 7E

The moles of ammonia that can be oxidized by $\text{268}\text{g}$ of oxygen are $6.70\text{mol}$.

Explanation of Solution

The balanced equation for the reaction between ammonia and oxygen are given below.

${\text{4NH}}_{\text{3}}+{\text{5O}}_{\text{2}}\to \text{4NO}+{\text{6H}}_{\text{2}}\text{O}$

Therefore, $4$ moles of ${\text{NH}}_{\text{3}}$ react with $\text{5}$ moles of ${\text{O}}_{\text{2}}$.

Therefore mole to mole ratio are given below.

$4\text{mol}{\text{NH}}_{\text{3}}=5\text{mol}{\text{O}}_{\text{2}}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{4\text{mol}{\text{NH}}_{\text{3}}}{5\text{mol}{\text{O}}_{\text{2}}}\text{and}\frac{5\text{mol}{\text{O}}_{\text{2}}}{4\text{mol}{\text{NH}}_{\text{3}}}$

The conversion factor to obtain moles of ${\text{NH}}_{\text{3}}$ from ${\text{O}}_{\text{2}}$ is given below.

$\frac{4\text{mol}{\text{NH}}_{\text{3}}}{5\text{mol}{\text{O}}_{\text{2}}}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{O}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 2\times 16.00\text{g}{\text{mol}}^{-1}\\ & =& \text{32}\text{.00}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to determine moles of ${\text{O}}_{\text{2}}$ from grams of ${\text{O}}_{\text{2}}$ is given below.

$\frac{1\text{mol}{\text{O}}_{\text{2}}}{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}$

The formula to calculate the number of moles of ${\text{NH}}_{\text{3}}$ from grams of ${\text{O}}_{\text{2}}$ is given below.

$\text{Moles}\text{of}{\text{NH}}_3=\left(\begin{array}{l}\text{Given}\text{mass}\text{of}{\text{O}}_{\text{2}}\times \text{Conversion}\text{factor}\text{to}\text{determine moles of}{\text{O}}_{\text{2}}\\ \times \text{Conversion}\text{factor}\text{to obtain moles of}{\text{NH}}_3 \end{array}\right)$…(1)

The mass of ${\text{O}}_{\text{2}}$ is $\text{268}\text{g}$.

Substitute the mass of ${\text{O}}_{\text{2}}$ and the conversion factors in equation (1).

$\begin{array}{rll}\text{Moles}\text{of}{\text{NH}}_3& =& \text{268}\text{g}{\text{O}}_{\text{2}}\times \frac{1\text{mol}{\text{O}}_{\text{2}}}{\text{32}\text{.00}\text{g}{\text{O}}_{\text{2}}}\times \frac{4\text{mol}{\text{NH}}_{\text{3}}}{5\text{mol}{\text{O}}_{\text{2}}}\\ & =& 6.70\text{mol}\end{array}$

Therefore, the moles of ${\text{NH}}_{\text{3}}$ that can be oxidized by $\text{268}\text{g}$ of oxygen are $6.70\text{mol}$.

Conclusion

The moles of ammonia that can be oxidized by $\text{268}\text{g}$ of oxygen are $6.70\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The grams of water will be produced if the reaction consumes $31.7$ moles of ammonia are to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 7E

The grams of water will be produced if the reaction consumes $31.7$ moles of ammonia are $\text{857}\text{g}$.

Explanation of Solution

The balanced equation for the reaction between ammonia and oxygen are given below.

${\text{4NH}}_{\text{3}}+{\text{5O}}_{\text{2}}\to \text{4NO}+{\text{6H}}_{\text{2}}\text{O}$

Therefore, $4$ moles of ${\text{NH}}_{\text{3}}$ react to give $\text{6}$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Therefore mole to mole ratio are given below.

$4\text{mol}{\text{NH}}_{\text{3}}=6\text{mol}{\text{H}}_{\text{2}}\text{O}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{4\text{mol}{\text{NH}}_{\text{3}}}{6\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{6\text{mol}{\text{H}}_{\text{2}}\text{O}}{4\text{mol}{\text{NH}}_{\text{3}}}$

The conversion factor to obtain moles of ${\text{H}}_{\text{2}}\text{O}$ from ${\text{NH}}_{\text{3}}$ is given below.

$\frac{6\text{mol}{\text{H}}_{\text{2}}\text{O}}{4\text{mol}{\text{NH}}_{\text{3}}}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.008}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{H}}_{\text{2}}\text{O}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times \text{1}\text{.008}\text{g}{\text{mol}}^{-1}\right)+\text{16}\text{.00}\text{g}{\text{mol}}^{-1}\\ & =& \text{18}\text{.016}\text{g}{\text{mol}}^{-1}\end{array}$

The conversion factor to calculate the grams of ${\text{H}}_{\text{2}}\text{O}$ is given below.

$\frac{\text{18}\text{.016}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{g}{\text{H}}_{\text{2}}\text{O}}$

The formula to calculate the number of grams of ${\text{H}}_{\text{2}}\text{O}$ from moles of ${\text{NH}}_{\text{3}}$ are given below.

$\text{Grams}\text{of}{\text{H}}_2\text{O}=\left(\begin{array}{l}\text{Given}\text{moles}\text{of}{\text{NH}}_3\times \text{Conversion}\text{factor}\text{to obtain moles of}{\text{H}}_2\text{O} \times \\ \text{Conversion}\text{factor}\text{to obtain}\text{grams}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}\right)$

The number of moles of ${\text{NH}}_{\text{3}}$ are $\text{31}\text{.7}\text{mol}$.

Substitute the moles of ${\text{NH}}_{\text{3}}$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Grams}\text{of}{\text{H}}_{\text{2}}\text{O}& =& \text{31}\text{.7}\text{mol}{\text{NH}}_{\text{3}}\times \frac{6\text{mol}{\text{H}}_{\text{2}}\text{O}}{4\text{mol}{\text{NH}}_{\text{3}}}\times \frac{\text{18}\text{.016}\text{g}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{g}{\text{H}}_{\text{2}}\text{O}}\\ & =& \text{856}\text{.67}\text{g}\\ & \simeq & \text{857}\text{g}\end{array}$

The grams of water will be produced if the reaction consumes $31.7$ moles of ammonia are $\text{857}\text{g}$.

Conclusion

The grams of water will be produced if the reaction consumes $31.7$ moles of ammonia are $\text{856}\text{.67}\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The grams of ammonia that are required to produce $\text{404}\text{g}$ of nitrogen monoxide are to be stated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 7E

The grams of ammonia that are required to produce $\text{404}\text{g}$ of nitrogen monoxide are $\text{299}\text{g}$.

Explanation of Solution

The balanced equation for the reaction between ammonia and oxygen is given below.

${\text{4NH}}_{\text{3}}+{\text{5O}}_{\text{2}}\to \text{4NO}+{\text{6H}}_{\text{2}}\text{O}$

Therefore, $4$ moles of ${\text{NH}}_{\text{3}}$ react to give $\text{4}$ moles of $\text{NO}$.

Therefore mole to mole ratio is given below.

$4\text{mol}{\text{NH}}_{\text{3}}=4\text{mol}\text{NO}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{4\text{mol}{\text{NH}}_{\text{3}}}{4\text{mol}\text{NO}}\text{and}\frac{4\text{mol}\text{NO}}{4\text{mol}{\text{NH}}_{\text{3}}}$

The conversion factor to obtain moles of $\text{NO}$ from ${\text{NH}}_{\text{3}}$ is given below.

$\frac{4\text{mol}{\text{NH}}_{\text{3}}}{4\text{mol}\text{NO}}$

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{NO}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 14.01\text{g}{\text{mol}}^{-1}+\text{16}\text{.00}\text{g}{\text{mol}}^{-1}\\ & =& \text{30}\text{.01}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain moles of $\text{NO}$ from grams of $\text{NO}$ is given below.

$\frac{1\text{mol}\text{NO}}{\text{30}\text{.01}\text{g}\text{NO}}$

The molar mass of hydrogen is $1.008\text{g}{\text{mol}}^{-1}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{NH}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 14.01\text{g}{\text{mol}}^{-1}+\left(3\times 1.008\text{g}{\text{mol}}^{-1}\right)\\ & =& \text{17}\text{.034}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain grams of ${\text{NH}}_{\text{3}}$ from moles of ${\text{NH}}_{\text{3}}$ is given below.

$\frac{\text{17}\text{.034}\text{g}{\text{NH}}_{\text{3}}}{1\text{mol}{\text{NH}}_{\text{3}}}$

The formula to calculate the grams ${\text{NH}}_{\text{3}}$ from the grams of $\text{NO}$ is given below.

$\text{Grams}\text{of}{\text{NH}}_{\text{3}}=\left(\begin{array}{l}\text{Given}\text{grams}\text{of}\text{NO}\times \text{Conversion}\text{factor}\text{to obtain moles of}\text{NO}\\ \times \text{Conversion}\text{factor}\text{to obtain grams}\text{of}{\text{NH}}_{\text{3}}\\ \times \text{Conversion}\text{factor}\text{to obtain}\text{moles}\text{of}{\text{NH}}_{\text{3}}\end{array}\right)$

The grams of $\text{NO}$ is $\text{404}\text{g}$.

Substitute the grams of $\text{NO}$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Grams}\text{of}{\text{NH}}_3& =& \text{404}\text{g}\text{NO}\times \frac{4\text{mol}{\text{NH}}_{\text{3}}}{4\text{mol}\text{NO}}\times \frac{1\text{mol}\text{NO}}{\text{30}\text{.01}\text{g}\text{NO}}\times \frac{\text{17}\text{.034}\text{g}{\text{NH}}_{\text{3}}}{1\text{mol}{\text{NH}}_{\text{3}}}\\ & =& \text{299}\text{g}\end{array}$

Therefore, the grams of ammonia that are required to produce $\text{404}\text{g}$ of nitrogen monoxide are $\text{299}\text{g}$.

Conclusion

The grams of ammonia that are required to produce $\text{404}\text{g}$ of nitrogen monoxide are $\text{299}\text{g}$.

Interpretation Introduction

(d)

Interpretation:

The yield of nitrogen monoxide (in grams) is to be stated if $\text{6}\text{.41}\text{g}$ of water is produced by the reaction.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in an equation which determines the moles of the reactants and products in the reaction.

Answer to Problem 7E

The yield of nitrogen monoxide (in grams) is $\text{7}\text{.12}\text{g}$.

Explanation of Solution

The balanced equation for the reaction between ammonia and oxygen are given below.

${\text{4NH}}_{\text{3}}+{\text{5O}}_{\text{2}}\to \text{4NO}+{\text{6H}}_{\text{2}}\text{O}$

Therefore, $4$ moles of $\text{NO}$ react with $\text{6}$ moles of ${\text{H}}_{\text{2}}\text{O}$.

Therefore mole to mole ratio are given below.

$4\text{mol}\text{NO}=6\text{mol}{\text{H}}_{\text{2}}\text{O}$

Therefore, two conversion factors from the mole-to-mole ratio are given below.

$\frac{4\text{mol}\text{NO}}{6\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{and}\frac{6\text{mol}{\text{H}}_{\text{2}}\text{O}}{4\text{mol}\text{NO}}$

The conversion factor to obtain moles of $\text{NO}$ from ${\text{H}}_{\text{2}}\text{O}$ is given below.

$\frac{4\text{mol}\text{NO}}{6\text{mol}{\text{H}}_{\text{2}}\text{O}}$

The molar mass of oxygen is $\text{16}\text{.00}\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.008}\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of ${\text{H}}_{\text{2}}\text{O}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \left(2\times \text{1}\text{.008}\text{g}{\text{mol}}^{-1}\right)+\text{16}\text{.00}\text{g}{\text{mol}}^{-1}\\ & =& \text{18}\text{.016}\text{g}{\text{mol}}^{-1}\end{array}$

The conversion factor to calculate the moles of ${\text{H}}_{\text{2}}\text{O}$ from grams of ${\text{H}}_{\text{2}}\text{O}$ is given below.

$\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.016}\text{g}{\text{H}}_{\text{2}}\text{O}}$

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

Therefore, the molar mass of $\text{NO}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 14.01\text{g}{\text{mol}}^{-1}+\text{16}\text{.00}\text{g}{\text{mol}}^{-1}\\ & =& \text{30}\text{.01}\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the conversion factor to obtain grams of $\text{NO}$ from moles of $\text{NO}$ is given below.

$\frac{\text{30}\text{.01}\text{g}\text{NO}}{1\text{mol}\text{NO}}$

The formula to calculate the grams of $\text{NO}$ from grams of ${\text{H}}_{\text{2}}\text{O}$ is given below.

$\text{Grams}\text{of}\text{NO}=\left(\begin{array}{l}\text{Given}\text{grams}\text{of}{\text{H}}_2\text{O}\times \text{Conversion}\text{factor}\text{to obtain moles of}{\text{H}}_2\text{O}\\ \times \text{Conversion}\text{factor}\text{to obtain}\text{grams}\text{of}\text{NO}\\ \times \text{conversion factor to obtain moles of }\text{NO}\text{from}{\text{H}}_2\text{O} \end{array}\right)$

The grams of ${\text{H}}_{\text{2}}\text{O}$ are $\text{6}\text{.41}\text{g}$.

Substitute the grams of ${\text{H}}_{\text{2}}\text{O}$ and the conversion factors in the above equation.

$\begin{array}{rll}\text{Grams}\text{of}\text{NO}& =& \text{6}\text{.41}{\text{H}}_{\text{2}}\text{O}\times \frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.016}\text{g}{\text{H}}_{\text{2}}\text{O}}\times \frac{\text{30}\text{.01}\text{g}\text{NO}}{1\text{mol}\text{NO}}\times \frac{4\text{mol}\text{NO}}{6\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ & =& \text{7}\text{.118}\text{g}\\ & \simeq & \text{7}\text{.12}\text{g}\end{array}$

Therefore, the yield of nitrogen monoxide (in grams) is $\text{7}\text{.12}\text{g}$.

Conclusion

The yield of nitrogen monoxide (in grams) is $\text{7}\text{.12}\text{g}$.