Chapter 13, Problem 75A

(a)

Interpretation Introduction

Interpretation:

On the basis of the given information, the number of moles of nitrogen gas and water is to be determined

Concept Introduction :

The ideal gases are those gases which obey the ideal gas equation and the ideal gas equation is written as

  $\text{PV}\text{= nRT}$

Where, P = pressure of the gas

V = Volume of the gas

R = Gas constant

T = temperature of the gas

Answer to Problem 75A

In the given reaction, $\text{0}\text{.295}\text{moles}$ of nitrogen gas and $\text{0}\text{.75}\text{moles}$ of water are produced.

Explanation of Solution

Given information:

The given unbalanced equation is:

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

Now, the balanced equation is:

  ${\text{4NH}}_{\text{3}}\left(\text{g}\right)\text{+}{\text{3O}}_{\text{2}}\left(\text{g}\right)\to {\text{2N}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Mass of ammonia gas = $\text{10}\text{.0g}$

Mass of oxygen gas = $\text{12}\text{.0g}$

The Molar mass of ammonia gas = $\text{17}\text{.031}\text{g/mol}$

The Molar mass of oxygen gas = $\text{31}\text{.999}\text{g/mol}$

Calculation:

To find the moles of nitrogen gas in ${\text{NH}}_{\text{3}}$

:

Mass of nitrogen gas

  $\text{0}\text{.587moles}\left({\text{NH}}_{\text{3}}\right)\frac{\text{2moles}{\text{N}}_{\text{2}}}{\text{4}\text{moles}{\text{NH}}_{\text{3}}}\left(\text{14}\text{.0067g/mol}\right){\text{N}}_{\text{2}}\text{=}\text{4}\text{.111g}$

The mass of nitrogen gas is, ${\text{m}}_{{\text{N}}_{\text{2}}}\text{=}\text{4}\text{.111g}$

Number of moles is,

  $\begin{array}{l}{\text{n}}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{4}\text{.111}}{\text{14}\text{.0667}}\\ \text{=}\text{0}\text{.2935}\text{mol}\end{array}$

To find the number of moles of water:

  $\text{0}\text{.375moles}\left({\text{O}}_{\text{2}}\right)\frac{\text{6moles}{\text{N}}_{\text{2}}}{\text{3}\text{moles}{\text{NH}}_{\text{3}}}\left(\text{18}\text{.02g/mol}\right){\text{H}}_{\text{2}}\text{O =}\text{13}\text{.515g}$

The mass of water is $\text{13}\text{.515g}$ .

  $\begin{array}{l}{\text{n}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\frac{\text{13}\text{.515}}{\text{18}\text{.02}}\\ \text{=}\text{0}\text{.75}\text{moles}\end{array}$

(b)

Interpretation Introduction

Interpretation: The total volume after completing a reaction is to be determined.

Concept Introduction :

The ideal gases are those gases which obey the ideal gas equation and the ideal gas equation is written as

  $\text{PV}\text{= nRT}$

Where, P = pressure of the gas

V = Volume of the gas

R = Gas constant

T = temperature of the gas

Answer to Problem 75A

After completing the reaction, the total volume is $\text{26}\text{.3L}$ .

Explanation of Solution

Given information:

Pressure = $\text{1atm}$

Temperature = $\text{35°C}\text{=}\text{35+273}\text{=}\text{308K}$

Calculated number of moles of nitrogen gas = $\text{0}\text{.295moles}$

Calculated number of moles of water = $\text{0}\text{.75moles}$

Calculation:

Ideal gas equation, $\text{PV=nRT}$

Volume of nitrogen gas

  $\begin{array}{l}{\text{V}}_{{\text{N}}_{\text{2}}}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{0}\text{.2935×0}\text{.0821×308}}{\text{1}}\\ \text{=}\text{7}\text{.4}\text{L}\end{array}$

And, the volume of water

  $\begin{array}{l}{\text{V}}_{{\text{H}}_{\text{2}}\text{O}}\text{=}\frac{{\text{n}}_{{\text{H}}_{\text{2}}\text{O}}\text{RT}}{\text{P}}\\ \text{=}\frac{\text{0}\text{.75×0}\text{.0821×308}}{\text{1}}\\ \text{=}\text{18}\text{.9}\text{L}\end{array}$

The total volume is

  $\begin{array}{l}{\text{V}}_{\text{T}}\text{=}{\text{V}}_{{\text{N}}_{\text{2}}}\text{+}{\text{V}}_{{\text{H}}_{\text{2}}\text{O}}\\ \text{=}\text{7}\text{.4}\text{+}\text{18}\text{.9}\\ \text{=}\text{26}\text{.3L}\end{array}$