Chapter 13, Problem 60A

Interpretation Introduction

Interpretation:

The volume of the oxygen gas required converting 5g sulfur to sulfur trioxide in the given reaction under given value of pressure and temperature is to be calculated.

Concept introduction:

1. 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

2. The substance having the number of moles which is given as

  $\text{number}\text{of}\text{moles}\text{=}\frac{\text{given}\text{mass}}{\text{molar}\text{mass}}$

Explanation of Solution

Given information:

Temperature = $\text{350°C}\text{=}\text{350+273}\text{=}\text{623K}$

Pressure = $\text{5}\text{.25}\text{atm}$

Mass of sulfur = $\text{5}\text{.00g}$

Calculation:

To find the moles of sulfur:

The given reaction is

  $\text{S}\left(\text{s}\right){\text{+O}}_{\text{2}}\left(\text{g}\right)\to {\text{SO}}_{\text{2}}\left(\text{g}\right)$

  $\begin{array}{l}{\text{n}}_{\text{S}}\text{=}\text{5g}\left(\text{S}\right)\frac{\text{1mole}}{\text{32}\text{.07}\text{g}\left(\text{s}\right)}\\ {\text{n}}_{\text{S}}\text{=}\text{0}\text{.156}\text{moles}\end{array}$

In the given reaction, 1 mole of ${\text{O}}_{\text{2}}$ gas reacts with 1 mole of sulfur

According to given values, number of moles of oxygen gas

  $\begin{array}{l}{\text{n}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.156}\left(\text{S}\right)\frac{\text{1}\text{mole}{\text{O}}_{\text{2}}}{\text{1}\text{mole}\text{S}}\\ {\text{n}}_{{\text{O}}_{\text{2}}}\text{=0}\text{.156}\text{moles}\end{array}$

The balanced chemical equation is

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

The number of moles of sulfur in ${\text{SO}}_{\text{2}}$ is $\text{0}\text{.156}\text{mole}$ .

Now, the number of moles of oxygen gas in this reaction is

  $\begin{array}{l}{\text{n}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.156moles}\left(\text{S}\right)\frac{\text{1}\text{mole}\left({\text{O}}_{\text{2}}\right)}{\text{2}\text{mole}\left(\text{S}\right)}\\ \text{=}\text{0}\text{.078}\text{moles}\end{array}$

The volume of oxygen gas at a given temperature and pressure is calculated as:

  $\begin{array}{l}\text{According to ideal gas law,}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{0}\text{.078×0}\text{.0821×623}}{\text{5}\text{.25}}\\ \text{=}\frac{\text{3}\text{.9895}}{\text{5}\text{.25}}\\ \text{=}\text{0}\text{.76}\text{L}\end{array}$

Conclusion

Under the given temperature and pressure, the volume of oxygen gas is $\text{0}\text{.76}\text{L}$