Chapter 6, Problem 143QP

(a)

Interpretation Introduction

Interpretation:

The reactant in excess and the amount of it remain unreacted are to be determined when $10.0\text{ mol}$ of ${\text{N}}_2$ is mixed with $10.0\text{ mol}$ of ${\text{H}}_2$ . The amount of reactant that would be mixed to make the statement true is to be determined.

Explanation of Solution

One mole of ${\text{N}}_2$ reacts with three moles of ${\text{H}}_2$ . Thus, ${\text{N}}_2$ is present in excess. So, the number of moles of ${\text{N}}_2$ that react with ${\text{H}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Number of moles of N}}_2=\frac{1{\text{ mol N}}_2}{3{\text{ mol H}}_2}\times 10{\text{ mol H}}_2\\ =3.33\end{array}$

This means that the reaction of $10.0\text{ mol}$ of ${\text{H}}_2$ with $3.33\text{ mol}$ of ${\text{N}}_2$ takes place. Since the number of moles of ${\text{H}}_2$ are limiting, the number of moles of ${\text{N}}_2$ are in excess. The number of unreacted moles of ${\text{N}}_2$ are calculated as follows:

$\begin{array}{c}\text{Number of unreacted moles}=\text{Total moles}-\text{Moles reacted}\\ =\text{10}\text{.0 mol}-3.33\text{ mol}\\ =\text{6}\text{.7 mol}\end{array}$

Since ${\text{H}}_2$ is in excess, the number of moles of ${\text{H}}_2$ that are mixed with ${\text{N}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of H}}_{\text{2}}\text{ reacted}=\text{Total moles}-\text{Unreacted moles}\\ =\text{10}\text{.0 mol}-1.0\text{ mol}\\ =\text{9}\text{.0 mol}\end{array}$

Thus, $9.0\text{ mol}$ of ${\text{H}}_2$ reacts with $3.0\text{ mol}$ of ${\text{N}}_2$ , so one mole of ${\text{H}}_2$ will remain unreacted.

(b)

Interpretation Introduction

Interpretation:

The amount of reactant that would need to be mixed to make the statement true is to be determined.

Explanation of Solution

According to the statement, ${\text{H}}_2$ is in excess, the number of moles of ${\text{H}}_2$ that are mixed with ${\text{N}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of H}}_{\text{2}}\text{ reacted}=\text{Total moles}-\text{Unreacted moles}\\ =\text{10}\text{.0 mol}-7.0\text{ mol}\\ =\text{3}\text{.0 mol}\end{array}$

${\text{N}}_2$ is also in excess, so to make ${\text{H}}_2$ in excess, the number of moles of ${\text{N}}_2$ required in the reaction should be decreased to $1$ . Thus, $1.0\text{ mol}$ of ${\text{N}}_2$ reacts with $3.00\text{ mol}$ of ${\text{H}}_2$ , so that seven moles of ${\text{H}}_2$ remain unreacted.

(c)

Interpretation Introduction

Interpretation:

The amount of reactant that would need to be mixed to make the statement true is to be determined.

Explanation of Solution

According to the statement, ${\text{N}}_2$ is in excess, the number of moles of ${\text{N}}_2$ that are mixed with ${\text{H}}_2$ are calculated as follows:

$\begin{array}{c}{\text{Moles of N}}_{\text{2}}\text{ reacted}=\text{Total moles}-\text{Unreacted moles}\\ =\text{10}\text{.0 mol}-1.0\text{ mol}\\ =\text{9}\text{.0 mol}\end{array}$

Thus, $9.0\text{ mol}$ of ${\text{N}}_2$ reacts with $27.0\text{ mol}$ of ${\text{H}}_2$ .

(d)

Interpretation Introduction

Interpretation:

The amount of reactant that would need to be mixed to make the statement true is to be determined.

Explanation of Solution

The number of unreacted moles of ${\text{N}}_2$ are calculated as follows:

$\begin{array}{c}\text{Number of unreacted moles}=\text{Total moles}-\text{Moles reacted}\\ =\text{10}\text{.0 mol}-3.33\text{ mol}\\ =\text{6}\text{.7 mol}\end{array}$

Thus, the statement is true.

(e)

Interpretation Introduction

Interpretation:

The amount of reactant that would need to be mixed to make the statement true is to be determined.

Explanation of Solution

${\text{N}}_2$ is in excess, so the limiting reactant should be ${\text{H}}_2$ .