Chapter 20, Problem 75A

(a)

Interpretation Introduction

Interpretation: The given redox reaction, $CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+C{O}_2\left(g\right)$ is to be balanced.

Concept Introduction: A reaction will be balanced when all the species involved in the given reaction are equal on both sides. That is in the reactant and product side.

Answer to Problem 75A

The balanced chemical equation is $5CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+5C{O}_2\left(g\right)$ .

Explanation of Solution

The given reaction is a redox reaction,

  $CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+C{O}_2\left(g\right)$

By carefully observing the reaction the number of carbon and iodine atoms is equal on both sides. But the number of oxygen is not equal. It can be balanced by adding 5 to CO and 5 to $C{O}_2$ . Thus, the equation becomes,

  $5CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+5C{O}_2\left(g\right)$

(b)

Interpretation Introduction

Interpretation: The oxidized and reduced species in the given reaction $CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+C{O}_2\left(g\right)$ is to be determined.

Concept Introduction: The substance that involves the loss of electrons are oxidized species. And the substance that involves a gain of electrons are reduced species. For oxidized species, the oxidation number will increase and for reduced species, the oxidation number will decrease.

Answer to Problem 75A

The oxidized species in the given reaction is C in CO and the reduced species is I in $I_2{O}_5$ .

Explanation of Solution

The given reaction,

  $CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+C{O}_2\left(g\right)$

The oxidation is happening to C in CO because its oxidation number is increasing from +2 to +4 in $C{O}_2$ . While the reduction is happening to I in $I_2{O}_5$ since the oxidation number is decreasing from +5 to 0. So oxidized is C in CO and reduced is I in $I_2{O}_5$ .

(c)

Interpretation Introduction

Interpretation: The amount of CO removed by 0.55g of $I_2{O}_5$ in the given redox reaction, $CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+C{O}_2\left(g\right)$ is to be determined.

Concept Introduction: The reactions which involve both oxidation and reduction are redox reactions.

The expression for moles is as follows:

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

Answer to Problem 75A

  $0.55g$ of $I_2{O}_5$ can remove $0.23g$ of CO in the air.

Explanation of Solution

The balanced chemical equation is,

  $5CO\left(g\right)+I_2{O}_5\left(s\right)\to I_2\left(s\right)+5C{O}_2\left(g\right)$

Thus one mole of $I_2{O}_5$ can remove 5CO molecules.

Substitute, $0.55$ for the mass of iodine and $333\text{g/mol}$ for molar mass of in equation (1).

  $\begin{array}{c}\text{moles}=\frac{\text{ 0}\text{.55}}{\text{333 g/mol}}\\ =0.0016\text{ mol}\end{array}$

That is the number of moles of iodine is $0.0016\text{ mol}$ . Therefore it can remove,

  $0.0016\text{ mol}\times \text{5}=0.008mol$ of CO in the atmosphere.

Hence,

Substitute, $0.008\text{ mol}$ for moles of iodine and $28$ for molar mass of in equation (1).

  $\begin{array}{c}\text{mass}\text{of solute}=\text{28}\times \text{0}\text{.008}\\ =0.23g\end{array}$

Thus $0.55g$ of $I_2{O}_5$ can remove $0.23g$ of CO.