Chapter 20, Problem 74A

Interpretation Introduction

Interpretation: The milliliters of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ of 0.28M required to oxidize $1.4\text{g}$ of sulfur of the given reaction, ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}+{\text{H}}_{\text{2}}\text{O}+\text{S}\to {\text{SO}}_{\text{2}}+\text{KOH}+{\text{Cr}}_{\text{2}}{\text{O}}_3$ is to be determined.

Concept introduction: The process which involves both oxidation and reduction reactions is defined as redox reactions.

The expression for moles is as follows:

  $\text{moles}=\frac{\text{ mass of solute}}{\text{Molar mass of solute}}$   ...... (1)

The formula to calculate the molarity of the solution is given as

  $\text{Molarity}=\frac{\text{Moles solute}}{\text{Volume }\left(\text{Litre}\right)\text{ solution}}$   ...... (2)

Answer to Problem 74A

  $104\text{mL}$ of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is needed to oxidize $1.4\text{g}$ of S.

Explanation of Solution

The reaction,

  ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}+{\text{H}}_{\text{2}}\text{O}+\text{S}\to {\text{SO}}_{\text{2}}+\text{KOH}+{\text{Cr}}_{\text{2}}{\text{O}}_3$

It can be balanced by the addition of a proper amount of numbers as coefficients.

The balanced equation of the given equation is,

  ${\text{2K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}+2{\text{H}}_{\text{2}}\text{O}+3\text{S}\to 3{\text{SO}}_{\text{2}}+4\text{KOH}+2{\text{Cr}}_{\text{2}}{\text{O}}_3$

The number of moles of S can be calculated from its given mass.

Substitute, $1.4\text{g}$ for the mass of $\text{S}$ and $32\text{g/mol}$ for molar mass in equation (1).

  $\begin{array}{c}\text{moles}=\frac{\text{ 1}\text{.4 g}}{\text{32 g/mol}}\\ =0.04375\text{ mol}\end{array}$

From the balanced equation for oxidizing 3 moles of S, about 2 moles of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is required. So, one mole of S can be oxidized by about $\frac{2}{3}$ moles of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ . Thus, for $0.04375\text{ mol}$ of S needs,

  $0.04375\text{ mol}\times \frac{2}{3}molesof{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ .

That is,

  $0.029\text{ mol}$ of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is needed.

Substitute, $0.029\text{ mol}$ for moles of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ and $0.28$ for molarity of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ in equation (2).

  $\begin{array}{c}{\text{Volume of K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\frac{\text{0}\text{.029}}{0.28}\times \\ =0.0104L\\ =104mL\end{array}$

Thus $104mL$ of ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$ is needed.