Chapter 10.3, Problem 41PP

Interpretation Introduction

(a)

Interpretation:

The compound ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is to be identified as ionic or molecular compound. The number of moles of $\text{2}\text{.50}\text{kg}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is to be calculated.

Concept introduction:

The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of chemical reaction, the limiting reagent is fully utilized. The number of moles is calculated as by the formula shown below.

$\text{No}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 41PP

The compound ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is identified as an ionic compound. The number of moles of $\text{2}\text{.50}\text{kg}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $0.156\times {10}^2\text{mol}$.

Explanation of Solution

Let’s assume a bond is formed between two elements A and B. For a bond to be ionic, the electro negativity difference (?e) between A and B should be greater than 1.7.

The electro negativity of iron (Fe) is 1.83 and the electro negativity of oxygen is (O) is 3.44. The electro negativity difference (?e) between Fe and O is 1.61 which is very close to the ionic bond. Therefore, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is a polar covalent compound with ionic character.

The mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{2}\text{.50}\text{kg}$. The conversion of mass from kg to g is shown below.

$\begin{array}{c}\text{Mass}=\text{2}\text{.50}\text{kg}\\ =\text{2}\text{.50}\text{kg}\left(\frac{1000\text{g}}{1\text{kg}}\right)\\ \text{Mass}=2.5\times {10}^3\text{g}\end{array}$

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $159.\text{69}\text{g/mol}$. The number of moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated as shown below.

$\begin{array}{c}\text{No}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{2.50\times {10}^3\text{g}}{159.\text{69}\text{g/mol}}\\ =0.0156\times {10}^3\text{mol}\\ =0.156\times {10}^2\text{mol}\end{array}$

Therefore, the number of moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $0.156\times {10}^2\text{mol}$.

Interpretation Introduction

(b)

Interpretation:

The compound ${\text{PbCl}}_{\text{4}}$ is to be identified as an ionic or molecular compound. The number of moles of $\text{25}\text{.4}\text{mg}$ of ${\text{PbCl}}_{\text{4}}$ is to be calculated.

Concept introduction:

The reagent in reaction that controls the amount of product formed is termed as limiting reagent. After completion of the chemical reaction, the limiting reagent is fully utilized. The number of moles is calculated as by the formula shown below.

$\text{No}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 41PP

The ${\text{PbCl}}_{\text{4}}$ is identified as a molecular compound. The number of moles of $\text{25}\text{.4}\text{mg}$ of ${\text{PbCl}}_{\text{4}}$ is $7.3\times {10}^{-5}\text{mol}$.

Explanation of Solution

In ${\text{PbCl}}_{\text{4}}$, the lead (Pb) has +4 charge. Due to the presence of a higher positive charge $\left({\text{Pb}}^{\text{+4}}\right)$, it polarizes the electron cloud of chlorine towards it self as per Fajan’s rule. Due to this reason, the ${\text{PbCl}}_{\text{4}}$ becomes covalent. Therefore, ${\text{PbCl}}_{\text{4}}$ is identified as a molecular compound.

The mass of ${\text{PbCl}}_{\text{4}}$ is $\text{25}\text{.4}\text{mg}$. The conversion of mass from mg to g is shown below.

$\begin{array}{c}\text{Mass}=\text{25}\text{.4}\text{mg}\\ =\text{25}\text{.4}\text{mg}\left(\frac{1\text{g}}{1000\text{mg}}\right)\\ \text{Mass}=25.4\times {10}^{-3}\text{g}\end{array}$

The molar mass of ${\text{PbCl}}_{\text{4}}$ is $349.012\text{g/mol}$. The number of moles of ${\text{PbCl}}_{\text{4}}$ is calculated as shown below.

$\begin{array}{c}\text{No}\text{.}\text{of}\text{moles}\left(n\right)=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{25.4\times {10}^{-3}\text{g}}{349.012\text{g/mol}}\\ =0.073\times {10}^{-3}\text{mol}\\ =7.3\times {10}^{-5}\text{mol}\end{array}$

Therefore, the number of moles of ${\text{PbCl}}_{\text{4}}$ is $7.3\times {10}^{-5}\text{mol}$.