Chapter 9, Problem 115AP

Interpretation Introduction

Interpretation:

A balanced equation for the given hypothetical reaction is to be written, and the oxidation and hybridization state of sulfur in ${\text{S}}_{\text{8}},{\text{SO}}_{\text{2}},{\text{ and SO}}_{\text{3}}$ is to be determined, and the mass of ${\text{SO}}_{\text{3}}$

required and the mass of ${\text{SO}}_{\text{2}}$

produced is to be calculated.

Concept introduction:

The oxidation state of an element is zero. The sum of the oxidation states of all the elements is equal to zero in a molecule of a compound, and in case of an ion, it is equal to the charge on the ion. Oxygen has a fixed oxidation state of $-2$

in its compounds.

To find hybridisation of an atom in a molecule, at first draw the Lewis structure of the molecule. Find the number of electrons domains around the atom. This gives the number of hybrid orbitals required for bonding. The number of hybrid orbitals is equal to the number of atomic orbitals that hybridise. Thus, one $s$

and one $p$

orbital hybridize to form two $sp$ hybrid orbitals, one $s$

and two $p$

orbitals hybridize to form three $s{p}^2$

hybrid orbitals, and one $s$

and three $p$

orbitals hybridize to form four $s{p}^3$

hybrid orbitals.

The conversion factor is a fraction that is used to convert one unit to another. Use of more than one factor to find a solution is called dimensional analysis.

Answer to Problem 115AP

Solution: The balanced equation for the reaction is as follows:

${\text{S}}_{\text{8}}\left(s\right){\text{+16SO}}_{\text{3}}\left(g\right)\to 24{\text{SO}}_{\text{2}}\left(g\right)$

The oxidation state of sulfur in elemental sulfur ${\text{S}}_{\text{8}}$ is $0$; ${\text{SO}}_{\text{3}}$

is $+6$; ${\text{SO}}_{\text{2}}$

is $+4$. Hybridization state of ${\text{S}}_{\text{8}}$ is $s{p}^3$; ${\text{SO}}_{\text{3}}$

is $s{p}^{\text{2}}$; ${\text{SO}}_{\text{2}}$

is $s{p}^{\text{2}}$. Mass of ${\text{SO}}_{\text{3}}$

is $4.99\text{ kg }$

and mass of ${\text{SO}}_{\text{2}}$

produced is $5.99\text{ kg}$.

Explanation of Solution

The hypothetical reaction between elemental sulfur and sulfur trioxide is as follows:

${\text{S}}_{\text{8}}{\text{+SO}}_{\text{3}}\to {\text{SO}}_{\text{2}}$

The balanced equation for this reaction is as follows:

${\text{S}}_{\text{8}}\left(s\right){\text{+16SO}}_{\text{3}}\left(g\right)\to 24{\text{SO}}_{\text{2}}\left(g\right)$

In ${\text{SO}}_{\text{3}}$, the sum of the oxidation states of sulfur and three oxygen atoms is zero.

The oxidation state of oxygen is $-2$.

Let $x$

is the oxidation state of sulfur.

Then $x$

is calculated as follows:

$\begin{array}{c}x+\left(\left(-2\right)\times 3\right)=0\\ x=+6\end{array}$

The oxidation state of sulfur in ${\text{SO}}_{\text{3}}$

is $+6$

In ${\text{SO}}_{\text{2}}$, the sum of the oxidation states of sulfur and two oxygen atoms is zero.

Let $x$

is the oxidation state of sulfur and the oxidation state of oxygen is $-2$.

$x$

is calculated as follows:

$\begin{array}{c}\left(x+\left(\left(-2\right)\times 2\right)\right)=0\\ x=+4\end{array}$

The oxidation state of sulfur in ${\text{SO}}_{\text{2}}$

is $+4$.

The Lewis structure of ${\text{S}}_{\text{8}},{\text{SO}}_{\text{3}},{\text{ and SO}}_{\text{2}}$, respectively, is as follows:

In ${\text{S}}_{\text{8}}$, each sulfur atom is attached to two sulfur atoms and there are two lone pairs of electrons on each sulfur atom. Thus, electron domain is four representing four $s{p}^3$

hybrid orbitals.

In ${\text{SO}}_{\text{3}}$, the sulfur atom is attached to three oxygen atoms and there is no lone pair of electrons on the sulfur atom. Thus, electron domain is three, representing three $s{p}^2$

hybrid orbitals.

In ${\text{SO}}_{\text{2}}$, the sulfur atom is attached to two oxygen atoms and there is one lone pair of electrons on the sulfur atom. Thus, the electron domain is three representing three $s{p}^2$

hybrid orbitals.

Consider the balanced equation:

${\text{S}}_{\text{8}}\left(s\right){\text{+16SO}}_{\text{3}}\left(g\right)\to 24{\text{SO}}_{\text{2}}\left(g\right)$

One mole of ${\text{S}}_{\text{8}}$

combines with $16$

moles of ${\text{SO}}_{\text{3}}$ and forms $24$

moles of ${\text{SO}}_{\text{2}}$. Molar mass of ${\text{S}}_{\text{8}}$

is $256.56\text{g}/\text{mol}$ and the mass of ${\text{S}}_{\text{8}}$

is given as $\text{1 kg}$

$\left(\text{1kg}=\text{1000 g}\right)$.

Convert the mass of ${\text{S}}_{\text{8}}$

to moles as follows:

$\left(1.00\cancel{{\text{kg S}}_8}\right)\times \left(\frac{1000\cancel{{\text{g S}}_8}}{1\cancel{{\text{kg S}}_8}}\right)\left(\frac{1{\text{ mol S}}_8}{256.56\cancel{{\text{g S}}_8}}\right)=\text{3}{\text{.9 mol S}}_8$

One mole of ${\text{S}}_{\text{8}}$

combines with $16$

moles of ${\text{SO}}_{\text{3}}$.

Thus, for $\text{3}{\text{.9 mol S}}_8$, the number of moles of ${\text{SO}}_{\text{3}}$

needed is calculated as follows:

$\text{3}\text{.9 }\cancel{{\text{mol S}}_8}\times \frac{16{\text{ mol SO}}_3}{1\cancel{{\text{mol S}}_{\text{8}}}}=62.4{\text{ mol SO}}_3$

Molar mass of ${\text{SO}}_{\text{3}}$

is $80.06\text{g}/\text{mol}$.

Thus, the amount of ${\text{SO}}_3$ is calculated as follows:

$62.4\cancel{{\text{mol SO}}_3}\times \left(\frac{80.06\cancel{{\text{g SO}}_3}}{1\cancel{{\text{mol SO}}_3}}\right)\times \left(\frac{1{\text{ kg SO}}_3}{1000\cancel{{\text{g SO}}_3}}\right)=4.99{\text{ kg SO}}_3$

One mole of ${\text{S}}_{\text{8}}$

produces $24$

moles of ${\text{SO}}_{\text{2}}$

Thus, for $\text{3}{\text{.9 mol S}}_8$, the number of moles of ${\text{SO}}_{\text{2}}$

produced is calculated as follows:

$\text{3}\text{.9 }\cancel{{\text{mol S}}_8}\times \frac{24{\text{ mol SO}}_2}{1\cancel{{\text{mol S}}_{\text{8}}}}=93.6{\text{ mol SO}}_2$

Molar mass of ${\text{SO}}_{\text{2}}$

is $64.06\text{g}/\text{mol}$.

Thus, the amount of ${\text{SO}}_2$ is calculated as follows:

$93.6\cancel{{\text{mol SO}}_2}\times \left(\frac{64.06\cancel{{\text{g SO}}_2}}{1\cancel{{\text{mol SO}}_2}}\right)\times \left(\frac{1{\text{ kg SO}}_2}{1000\cancel{{\text{g SO}}_2}}\right)=5.99\text{}{\text{ kg SO}}_2$

Hence, mass of ${\text{SO}}_{\text{3}}$

required is $4.99\text{ kg }$ and mass of ${\text{SO}}_{\text{2}}$

produced is $5.99\text{ kg}$.

Conclusion

The balanced equation for the reaction between elemental sulfur and sulfur trioxide to produce sulfur dioxide is written; the oxidation and hybridization state of sulfur in ${\text{S}}_{\text{8}},{\text{SO}}_{\text{2}},{\text{ and SO}}_{\text{3}}$ are determined; and the mass of ${\text{SO}}_{\text{3}}$

required and the mass of ${\text{SO}}_{\text{2}}$

produced by $\text{1 kg}$

of elemental sulfur are calculated.