Chapter 11.3, Problem 23PP

Interpretation Introduction

Interpretation:

The limiting reactant needs to be determined for the given reaction when 100.0 g of $\text{Na}$ reacts with 100.0 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

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.

Answer to Problem 23PP

${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is the limiting reactant.

Explanation of Solution

Given:

The balanced reaction is:

$\text{6Na}\left(\text{s}\right){\text{+Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\to {\text{3Na}}_{\text{2}}\text{O}\left(\text{s}\right)\text{+2Fe}\left(\text{s}\right)$

It is given that 100.0 g of $\text{Na}$ reacts with 100.0 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

The molar mass of $\text{Na}$ and ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is 22.99 g/mol and 159.69 g/mol respectively. The number of moles of each is calculated using formula:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

So,

$\begin{array}{l}{\text{n}}_{\text{Na}}\text{= }\frac{\text{100}\text{.0 g}}{\text{22}\text{.99 g/mol}}\\ {\text{n}}_{\text{Na}}\text{= 4}\text{.35 mol}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{= }\frac{\text{100}\text{.0 g}}{\text{159}\text{.69 g/mol}}\\ {\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{= 0}\text{.63 mol}\end{array}$

In the given balanced reaction, 6.0 mol of $\text{Na}$ reacts with 1.0 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$. So, the number of moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ required to react with $\text{4}\text{.35 mol}$ of $\text{Na}$ is:

$4.35\text{ mol of Na×}\frac{{\text{1 mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{6 mol of Na}}\text{ = }0.725{\text{ mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

Since, the required moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $0.725\text{ mol}$ and available amount is $\text{0}\text{.63 mol}$ so, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is the limiting reactant.

Interpretation Introduction

Interpretation:

The reactant present in excess needs to be determined for the given reaction when 100.0 g of $\text{Na}$ reacts with 100.0 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

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.

Answer to Problem 23PP

$\text{Na}$ is present in excess.

Explanation of Solution

It is given that 100.0 g of $\text{Na}$ reacts with 100.0 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

The molar mass of $\text{Na}$ and ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is 22.99 g/mol and 159.69 g/mol respectively. The number of moles of each is calculated using formula:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

So,

$\begin{array}{l}{\text{n}}_{\text{Na}}\text{= }\frac{\text{100}\text{.0 g}}{\text{22}\text{.99 g/mol}}\\ {\text{n}}_{\text{Na}}\text{= 4}\text{.35 mol}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{= }\frac{\text{100}\text{.0 g}}{\text{159}\text{.69 g/mol}}\\ {\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{= 0}\text{.63 mol}\end{array}$

In the given balanced reaction, 6.0 mol of $\text{Na}$ reacts with 1.0 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$. So, the number of moles of $\text{Na}$ required to react with $\text{0}\text{.63 mol}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is:

$\text{0}{\text{.63 mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{6 mol of Na}}{{\text{1 mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{ = 3}\text{.78 mol of Na}$

Since, the required moles of $\text{Na}$ is $3.78\text{ mol}$ and available amount is $\text{4}\text{.35 mol}$ so, $\text{Na}$ is present in excess.

Interpretation Introduction

Interpretation:

The mass of solid iron produced needs to be calculated for the given reaction when 100.0 g of $\text{Na}$ reacts with 100.0 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

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.

Answer to Problem 23PP

The mass of solid iron produced is $\text{70}\text{.37 g}$.

Explanation of Solution

The balanced reaction is:

$\text{6Na}\left(\text{s}\right){\text{+Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\to {\text{3Na}}_{\text{2}}\text{O}\left(\text{s}\right)\text{+2Fe}\left(\text{s}\right)$

Since, $\text{Na}$ is present in excess so, the mass of $\text{Na}$ reacted with ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is calculated as:

From the balanced reaction, the mole ratio of $\text{Na}$ and ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is 6: 1 that means 1 mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ produces 2 moles of $\text{Fe}$ so, the number of moles of $\text{Fe}$ produced from 0.63 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is:

$0.63{\text{ mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{2 mol of Fe }}{{\text{1 mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{ = }1.26\text{ mol of Fe}$

Converting the moles of $\text{Fe}$ to mass as:

The molar mass of $\text{Fe}$ is 55.85 g/mol.

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

So,

$\begin{array}{l}\text{1}\text{.26 mol = }\frac{\text{mass}}{\text{55}\text{.85 g/mol}}\\ \text{mass = 1}\text{.26 mol}\times \text{55}\text{.85 g/mol}\\ \text{mass = 70}\text{.37 g}\end{array}$

Hence, the mass of solid iron produced is $\text{70}\text{.37 g}$.

Interpretation Introduction

Interpretation:

The mass excess reactant present after the completion of reaction needs to be calculated for the given reaction when 100.0 g of $\text{Na}$ reacts with 100.0 g of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

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.

Answer to Problem 23PP

The mass excess reactant present after the completion of reaction is $13.1\text{ g}$ of $\text{Na}$.

Explanation of Solution

The balanced reaction is:

$\text{6Na}\left(\text{s}\right){\text{+Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)\to {\text{3Na}}_{\text{2}}\text{O}\left(\text{s}\right)\text{+2Fe}\left(\text{s}\right)$

The mass of leftover reactant is calculated as:

The molar mass of $\text{Na}$ and ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is 22.99 g/mol and 159.69 g/mol respectively. The number of moles of each is calculated using formula:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

So,

$\begin{array}{l}{\text{n}}_{\text{Na}}\text{= }\frac{\text{100}\text{.0 g}}{\text{22}\text{.99 g/mol}}\\ {\text{n}}_{\text{Na}}\text{= 4}\text{.35 mol}\end{array}$

$\begin{array}{l}{\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{= }\frac{\text{100}\text{.0 g}}{\text{159}\text{.69 g/mol}}\\ {\text{n}}_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{= 0}\text{.63 mol}\end{array}$

In the given balanced reaction, 6.0 mol of $\text{Na}$ reacts with 1.0 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$. So, the number of moles of $\text{Na}$ required to react with $\text{0}\text{.63 mol}$ of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ (limiting reactant) is:

$\text{0}{\text{.63 mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{6 mol of Na}}{{\text{1 mol of Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\text{ = }3.78\text{ mol of Na}$

Thus, the leftover mole of $\text{Na}$ is:

$\text{4}\text{.35 mol - }3.78\text{ mol = 0}\text{.57 mol}$

Since, the molar mass of $\text{Na}$ is 22.99 g/mol so:

$\text{Number of moles}=\frac{\text{Mass}}{\text{Molar mass}}$

$\begin{array}{l}\text{0}\text{.57 mol}=\frac{\text{Mass}}{\text{22}\text{.99 g/mol}}\\ \text{Mass = 0}\text{.57 mol}\times \text{22}\text{.99 g/mol}\\ \text{Mass = }13.1\text{ g}\end{array}$

Hence, the mass excess reactant present after the completion of reaction is $13.1\text{ g}$ of $\text{Na}$.