Chapter 4, Problem 4.105QP

(a)

Interpretation Introduction

Interpretation:

The given reaction ${\text{N}}_{\text{2(g)}}{\text{ + H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(g)}}$ has to be balanced.

Concept Introduction:

Balancing the equation:

• There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• First write the skeletal reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Place suitable coefficients in front of reactants as well as products until the number of atoms on each side (reactants and products) becomes equal.

Explanation of Solution

Given equation:

  ${\text{N}}_{\text{2(g)}}{\text{ + H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(g)}}$

Balancing the chemical Equation:

Count the number of atoms on each side of the reaction.

Atom	Reactant side	Product side
$\text{N}$	2	1
$\text{H}$	2	3

Place suitable coefficient before reactants and products and then check for the number of atoms again.

  ${\text{N}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2NH}}_{\text{3(g)}}$

Atom	Reactant side	Product side
$\text{N}$	2	2
$\text{H}$	6	6

The number of atoms of present on each side of the reaction is same.  Hence, the balanced equation for the given reaction is ${\text{N}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2NH}}_{\text{3(g)}}$.

(b)

Interpretation Introduction

Interpretation:

For the given reaction, the number of ${\text{mol of H}}_{\text{2}}$ that would react with ${\text{1 mol of N}}_{\text{2}}$ has to be given.

Explanation of Solution

Given reaction:

  ${\text{N}}_{\text{2(g)}}{\text{ + H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(g)}}$

From subpart (a), the balanced equation of the given reaction is ${\text{N}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2NH}}_{\text{3(g)}}$.

From the stoichiometry of the balanced equation, three moles of ${\text{H}}_{\text{2}}$ reacts with one mole of ${\text{N}}_{\text{2}}$

(c)

Interpretation Introduction

Interpretation:

For the given reaction, the number of moles of product that would form from ${\text{1 mol of N}}_{\text{2}}$ has to be given.

Explanation of Solution

Given reaction:

  ${\text{N}}_{\text{2(g)}}{\text{ + H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(g)}}$

From subpart (a), the balanced equation of the given reaction is ${\text{N}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2NH}}_{\text{3(g)}}$.

From the stoichiometry of the balanced equation, one mole of ${\text{N}}_{\text{2}}$ reacts with hydrogen and forms two moles of ammonia as product.

(d)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{H}}_{\text{2}}$ that is required to react with $\text{14}\text{.0 g}$ of ${\text{N}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Moles:

Mole of the substance is found by dividing the mass of the substance by its molar mass.

  $\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Explanation of Solution

Given reaction:

  ${\text{N}}_{\text{2(g)}}{\text{ + H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(g)}}$

From subpart (a), the balanced equation of the given reaction is ${\text{N}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2NH}}_{\text{3(g)}}$.

Determination of moles:

From the balanced equation, it is known that one mole of ${\text{N}}_{\text{2}}$ reacts with three moles of ${\text{H}}_{\text{2}}$.  That is, ${\text{1 mol N}}_2{\text{ = 3 mol H}}_{\text{2}}$.

The molar mass of ${\text{N}}_{\text{2}}$ is $\text{28}\text{.013 g/mol}$.

Determine the moles of ${\text{N}}_{\text{2}}$ using given mass as follows,

  $14.0{\text{ g N}}_{\text{2}}\times \frac{{\text{ 1 mol N}}_{\text{2}}}{\text{ 28}{\text{.013 g N}}_{\text{2}}}=0.4998{\text{ mol of N}}_{\text{2}}$

One mole of ${\text{N}}_{\text{2}}$ reacts with three moles of ${\text{H}}_{\text{2}}$.  Therefore, using the molar ratio,

  $\begin{array}{l}0.4998{\text{ mol N}}_{\text{2}}\times \frac{{\text{3 mol H}}_{\text{2}}}{{\text{1 mol N}}_{\text{2}}}=\text{1}{\text{.4994 mol H}}_{\text{2}}\\ \approx 1.50{\text{ mol H}}_{\text{2}}\end{array}$

Thus, the number of moles of ${\text{H}}_{\text{2}}$ required is $1.50\text{ mol}$.

(e)

Interpretation Introduction

Interpretation:

The amount (in $\text{g}$) of product (ammonia) that is produced from $\text{14}\text{.0 g}$ of ${\text{N}}_{\text{2}}$ has to be calculated.

Concept Introduction:

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

  $\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Explanation of Solution

Given reaction:

  ${\text{N}}_{\text{2(g)}}{\text{ + H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ NH}}_{\text{3(g)}}$

From subpart (a), the balanced equation of the given reaction is ${\text{N}}_{\text{2(g)}}{\text{ + 3H}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2NH}}_{\text{3(g)}}$.

Determination of mass of product:

From the balanced equation, it is known that one mole of ${\text{N}}_{\text{2}}$ reacts with three moles of hydrogen and forms two moles of product, ${\text{NH}}_{\text{3}}$.  That is, ${\text{1 mol N}}_2{\text{ = 2 mol NH}}_3$.

The molar mass of ${\text{N}}_{\text{2}}$ is $\text{28}\text{.013 g/mol}$.

Determine the moles of ${\text{N}}_{\text{2}}$ using given mass as follows,

  $14.0{\text{ g N}}_{\text{2}}\times \frac{{\text{ 1 mol N}}_{\text{2}}}{\text{ 28}{\text{.013 g N}}_{\text{2}}}=0.4998{\text{ mol of N}}_{\text{2}}$

One mole of ${\text{N}}_{\text{2}}$ reacts and gives two moles of ${\text{NH}}_{\text{3}}$.  Therefore, using the molar ratio,

  $\begin{array}{l}0.4998{\text{ mol N}}_{\text{2}}\times \frac{{\text{2 mol NH}}_{\text{3}}}{{\text{1 mol N}}_{\text{2}}}=0.9996{\text{ mol NH}}_{\text{3}}\\ \approx 1.00{\text{ mol NH}}_{\text{3}}\end{array}$

Thus, the number of moles of ${\text{NH}}_{\text{3}}$ is $1.00{\text{ mol NH}}_{\text{3}}$.

The molar mass of ${\text{NH}}_{\text{3}}$ is $\text{17}\text{.031 g/mol}$.

Now, calculate the mass of ${\text{NH}}_{\text{3}}$ produced from given amount of ${\text{N}}_{\text{2}}$ as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 17}\text{.031 g/mol}\times 1.00{\text{ mol NH}}_{\text{3}}\\ =17.03\text{ g}\\ \approx 17{\text{ g NH}}_{\text{3}}\end{array}$

Therefore, the mass of ${\text{NH}}_{\text{3}}$ produced is $17\text{ g}$