Chapter 10, Problem 1E

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 will react with $95.3$ moles of oxygen?

b) How many moles of nitrogen monoxide will result from the reaction of $2.89$ moles

of ammonia?

c) If $3.35$ moles of water is produced, how many moles of nitrogen monoxide will

also be produced?

Interpretation Introduction

(a)

Interpretation:

The number of moles of ammonia that will react with $95.3$ moles of oxygen is 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 1E

The number of moles of ammonia that will react with $95.3$ moles of oxygen is $76.2\text{mol}$.

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 with $\text{5}$ moles of ${\text{O}}_{\text{2}}$.

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

$\text{Moles}\text{of}{\text{NH}}_3=\text{Given}\text{moles}\text{of}{\text{O}}_{\text{2}}\times \text{Conversion}\text{factor}$ …(1)

The number of moles of ${\text{O}}_{\text{2}}$ are $\text{95}\text{.3}\text{moles}$.

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

$\begin{array}{rll}\text{Moles}\text{of}{\text{NH}}_3& =& \text{95}\text{.3}\text{mol}{\text{O}}_{\text{2}}\times \frac{4\text{mol}{\text{NH}}_{\text{3}}}{5\text{mol}{\text{O}}_{\text{2}}}\\ & =& 76.2\text{mol}\end{array}$

Therefore, the number of moles of ${\text{NH}}_{\text{3}}$ that will react with $95.3$ moles of oxygen is $76.2\text{mol}$.

Conclusion

The number of moles of ammonia that will react with $95.3$ moles of oxygen is $76.2\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The number of moles of nitrogen monoxide will result from the reaction of $2.89$ moles of ammonia is 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 1E

The number of moles of nitrogen monoxide will result from the reaction of $2.89$ moles of ammonia is $\text{2}\text{.89}\text{mol}$.

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.

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

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

$\text{Moles}\text{of}\text{NO}=\text{Given}\text{moles}\text{of}{\text{NH}}_3\times \text{Conversion}\text{factor}$ …(2)

The number of moles of ${\text{NH}}_{\text{3}}$ are $\text{2}\text{.89}\text{mol}$.

Substitute the moles of ${\text{NH}}_{\text{3}}$ and conversion factor in equation (2).

$\begin{array}{rll}\text{Moles}\text{of}{\text{NH}}_3& =& \text{2}\text{.89}\text{mol}{\text{NH}}_{\text{3}}\times \frac{4\text{mol}\text{NO}}{4\text{mol}{\text{NH}}_{\text{3}}}\\ & =& \text{2}\text{.89}\text{mol}\end{array}$

Therefore, the number of moles of nitrogen monoxide will result from the reaction of $2.89$ moles of ammonia is $\text{2}\text{.89}\text{mol}$.

Conclusion

The number of moles of nitrogen monoxide will result from the reaction of $2.89$ moles of ammonia is $\text{2}\text{.89}\text{mol}$.

Interpretation Introduction

(c)

Interpretation:

The number of moles of nitrogen monoxide will be produced when $3.35$ moles of water is produced is 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 1E

The number of moles of nitrogen monoxide will be produced when $3.35$ moles of water is produced is $\text{2}\text{.23}\text{mol}$.

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, $6$ moles of ${\text{H}}_{\text{2}}\text{O}$ react with $\text{4}$ 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.

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

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

$\text{Moles}\text{of}\text{NO}=\text{Given}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\times \text{Conversion}\text{factor}$ …(3)

The number of moles of ${\text{H}}_{\text{2}}\text{O}$ are $\text{3}\text{.35}\text{mol}$.

Substitute the moles of ${\text{H}}_{\text{2}}\text{O}$ and conversion factor in equation (3).

$\begin{array}{rll}\text{Moles}\text{of}\text{NO}& =& \text{3}\text{.35}\text{mol}{\text{H}}_{\text{2}}\text{O}\times \frac{4\text{mol}\text{NO}}{6\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ & =& \text{2}\text{.23}\text{mol}\end{array}$

Therefore, the number of moles of nitrogen monoxide will be produced when $3.35$ moles of water is produced is $\text{2}\text{.23}\text{mol}$.

Conclusion

The number of moles of nitrogen monoxide will be produced when $3.35$ moles of water is produced is $\text{2}\text{.23}\text{mol}$.