Q72P - Chemistry: Molecular Nature Of Matter And Change

Question

Lead can be prepared from galena [lead(II) sulfide] by first roasting the galena in oxygen gas to form lead(II) oxide and sulfur dioxide. Heating the metal oxide with more galena forms the molten metal and more sulfur dioxide. (a) Write a balanced equation for each step. 

(b) Write an overall balanced equation for the process. 

(c) How many metric tons of sulfur dioxide form for every metric ton of lead obtained?

Step-by-Step Solution

Answer

(a) The balanced equation of the first step:

$\mathbf{2}\mathbf{PbS}\left(s\right)\mathbf{+}{\mathbf{30}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{2}\mathbf{PbO}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)$

The balanced equation of the second step:

$\mathbf{2}\mathbf{PbO}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{PbO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{3}\mathbf{Pb}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(b) The balanced equation of the overall reaction is:

$\mathbf{2}\mathbf{PbS}\left(s\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathbf{Pb}\left(l\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)$

(c) The mass of  ${\mathbf{SO}}_{\mathbf{2}}$is 0.31 metric tons.

Step 1: Write a balanced chemical equation of the roasting of galena in oxygen gas

When galena$\left(\mathrm{PbS}\right)$ is roasted in presence of oxygen$\left({\mathrm{O}}_2\right)$lead(II) oxide$\left(\mathrm{PbO}\right)$ and sulfur dioxide$\left(S{O}_2\right)$ are formed. The balanced chemical equation is:

$\mathbf{2}\mathbf{PbS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{30}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{2}\mathbf{PbO}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\left(i\right)$

Step 2: Write a balanced chemical equation of the heating of lead(II) oxide with more galena

When lead(II) oxide$\left(\mathrm{PbO}\right)$is heated with excessgalena$\left(\mathrm{PbS}\right)$molten lead$\left(\mathrm{Pb}\right)$ and sulfur dioxide$\left({\mathrm{SO}}_2\right)$ are formed. The balanced chemical equation is:

$\mathbf{2}\mathbf{PbO}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathbf{PbS}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{3}\mathbf{Pb}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\left(\mathrm{ii}\right)$

Step 3: Write a balanced equation of the overall reaction

The balanced overall equation of the reaction is written by adding equations (i) and (ii).

$\mathbf{3}\mathbf{PbS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{3}\mathbf{Pb}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\left(\mathrm{iii}\right)$

Divide equation (iii) by coefficient 3,

$\mathbf{PbS}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{\to }\mathbf{Pb}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Step 4: Relation between mass and number of moles

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathrm{Number} \mathrm{of} \mathrm{moles}=\frac{\mathrm{mass}}{\mathrm{Molar} \mathrm{mass}}$

Step 5: Calculate the number of moles of the final product ( Pb )

The mass of$\mathbf{Pb}$ = 1 metric ton.

Since 1 metric ton is equal to${\mathbf{10}}^{\mathbf{6}}$ grams.

The molar mass of $\mathbf{Pb}$ = 207.2 g/mol.

The number of moles of is calculated as:

$$\begin{gathered} \mathrm{moles} \mathrm{of} \mathrm{Pb}=\frac{\mathrm{mass} \mathrm{of} \mathrm{Pb}}{\mathrm{Molar} \mathrm{mass} \mathrm{of} \mathrm{Pb}} \\ =\frac{{10}^6\mathrm{g}}{207.2\mathrm{g}/\mathrm{mol}} \\ =4826.25\mathrm{mol} \end{gathered}$$

Therefore, the number of moles of $\mathbf{Pb}$ is 4826.25 mol.

Step 6: Relation between number of moles of Pb   and   SO 2

In the given reaction, 1mol of ${\mathbf{SO}}_{\mathbf{2}}$is formed with 1 mol of $\mathbf{Pb}$

Thus,

$1\mathrm{mol} \mathrm{of} \mathrm{Pb}=1\mathrm{mol} \mathrm{of} {\mathrm{SO}}_2$

Step 7: Calculate the number of moles of SO 2

The number of moles of ${\mathbf{SO}}_{\mathbf{2}}$is:

$$\begin{gathered} \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Pb}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SO}}_{\mathbf{2}} \\ \mathbf{4826}\mathbf{.}\mathbf{25}\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Pb}\mathbf{=}\mathbf{4826}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{SO}}_{\mathbf{2}} \end{gathered}$$

Step 8: Calculate the mass of SO 2

The molar mass of ${\mathbf{SO}}_{\mathbf{2}}$= 64.066 g/mol.

The mass of ${\mathbf{SO}}_{\mathbf{2}}$ is calculated as:

$$\begin{gathered} \mathrm{mass} \mathrm{of} {\mathrm{SO}}_2=\mathrm{moles} \mathrm{of} {\mathrm{SO}}_2\times \mathrm{Molar} \mathrm{mass} \mathrm{of} {\mathrm{SO}}_2 \\ = 4826.25\mathrm{mol}\times 64.66\mathrm{g}/\mathrm{mol} \\ = 309198.53\mathrm{g} \\ = 0.309919853\mathrm{metriction} \end{gathered}$$

Therefore, the mass of ${\mathbf{SO}}_{\mathbf{2}}$ is approximately 0.31 metric tons