Chapter 13, Problem 61A

Interpretation Introduction

Interpretation:

The volume of the oxygen gas under given information is to be determined and the volume of the sulphur dioxide gas is also 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 61A

The volume of the oxygen gas based on given information is $\text{3L}$ and volume of sulphur dioxide gas is $\text{3}\text{.9L}$

Explanation of Solution

Given information:

The given chemical reaction is

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

Calculation:

The balanced chemical equation is

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

Temperature = $\text{27}\text{.5°C}\text{=}\text{27}\text{.5+273=}\text{300}\text{.5 K}$

Pressure = $\text{0}\text{.998}\text{atm}$

Mass of copper = $\text{25g}$

To calculate number of moles of copper in copper sulphide

  $\begin{array}{l}{n}_{C{u}_{2}S}=\frac{25}{159.16}\\ =0.1571mole\end{array}$

To find the number of moles of volume:

  $\begin{array}{l}\text{0}\text{.1571}\text{mole}\left({\text{Cu}}_{\text{2}}\text{S}\right)\frac{\text{3}\text{mole}{\text{o}}_{\text{2}}}{\text{2}\text{mole}\left({\text{Cu}}_{\text{2}}\text{S}\right)}\text{=}\text{0}\text{.1178}\text{moles}\\ {\text{n}}_{{\text{O}}_{\text{2}}}\text{=}\text{0}\text{.1178}\text{moles}\end{array}$

The volume of the oxygen gas

  $\begin{array}{l}\text{According to ideal gas}\\ {\text{V}}_{{\text{O}}_{\text{2}}}\text{=}\frac{\text{0}\text{.1178×300}\text{.5×0}\text{.0821}}{\text{0}\text{.998}}\\ \text{=}\text{2}\text{.9}\text{L}\\ \text{=3L}\left(\text{Approximate}\right)\end{array}$

To find the mass of the sulphur dioxide gas:

  $\text{0}\text{.1571}\text{}\text{moles}\left({\text{Cu}}_{\text{2}}\text{S}\right)\frac{\text{2}\text{mole}\left({\text{SO}}_{\text{2}}\right)}{\text{2}\text{mole}\left({\text{Cu}}_{\text{2}}\text{S}\right)}\left(\text{64}\text{.07g}\right)\left({\text{SO}}_{\text{2}}\right)\text{=10}\text{.06g}$

  ${\text{m}}_{{\text{SO}}_{\text{2}}}\text{=}\text{10}\text{.06g}$

The volume of the sulphur dioxide gas:

Number of moles of sulphur dioxide gas is

  $\begin{array}{l}{\text{n}}_{{\text{SO}}_{\text{2}}}\text{=}\frac{\text{10}\text{.6g}}{\text{64}\text{.066}}\\ \text{=}\text{0}\text{.1571}\text{moles}\end{array}$

Now, the volume of the sulphur dioxide gas is

  $\begin{array}{l}{\text{V}}_{{\text{SO}}_{\text{2}}}\text{=}\frac{\text{0}\text{.1571×300}\text{.5×0}\text{.0821}}{\text{0}\text{.998}}\\ \text{=}\text{3}\text{.9}\text{L}\\ \text{=}\text{4L}\left(\text{Approximate}\right)\end{array}$

Conclusion

On the basis of the given values, the volume of oxygen gas is $\text{3L}$ and volume of sulphur dioxide gas $\text{3}\text{.9L}$ .