Chapter 27, Problem 27.EE

(a)

Interpretation Introduction

Interpretation:

Oxidation state of Cobalt in the formula of ${\text{LaCoO}}_{\text{3}}$ has to be found.

Concept introduction:

Oxidation number is defined as degree of oxidation (ie. loss of an electrons) of an atom within the chemical compound.  There is slightly difference between oxidation state and oxidation number.  In oxidation state, there is electronegativity of an atom in a bond should be considered.  However, when describing oxidation number, electronegativity does not considered.

Answer to Problem 27.EE

The oxidation state of $\text{Co}$ in ${\text{LaCoO}}_{\text{3}}$ is found to be +3.

Explanation of Solution

Given information:

Oxidation state of Oxygen is -2.

Oxidation state of Lanthanide is +3.

To found: Oxidation state of Cobalt in the formula of ${\text{LaCoO}}_{\text{3}}$

${\text{LaCoO}}_{\text{3}}$

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

Where,

$\text{x}$ is Co metal

Thus, oxidation state o Co is found to be +3.

(b)

Interpretation Introduction

Interpretation:

The reaction between ${\text{LaCoO}}_{\text{3}}$ with ${\text{H}}_{\text{2}}$ has to be written.

Explanation of Solution

To write: the reaction of ${\text{LaCoO}}_{\text{3}}$ with ${\text{H}}_{\text{2}}$

${\text{LaCoO}}_{\text{3}}$ reacts with ${\text{H}}_{\text{2}}$ to gives ${\text{La}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{,Co}}_{\left(\text{s}\right)}{\text{and H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$.

Thus reaction is,

${\text{LaCoO}}_{\text{3}}{}_{\left(\text{s}\right)}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }\underset{\text{FM}\text{221}\text{.8378}}{\underbrace{\frac{\text{1}}{\text{2}}{\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+ Co}}_{\left(\text{s}\right)}}}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$

(c)

Interpretation Introduction

Interpretation:

The mass of the product when $\text{41}\text{.8724}{\text{mg of LaCoO}}_{\text{3}}$ reacts completely has to be calculated.

Answer to Problem 27.EE

Mass of the product is found to be $\text{37}\text{.78473 mg}$.

Explanation of Solution

To calculate: mass of the product when $\text{41}\text{.8724}{\text{mg of LaCoO}}_{\text{3}}$ reacts completely

$\text{245}\text{.8369g}{\text{of LaCoO}}_{\text{3}}$ would create $\text{221}\text{.8378 g}$ of solid product

$\begin{array}{l}\frac{\text{Product}\text{from41}\text{.8724mg}}{\text{41}\text{.8724}\text{mg}}\text{= }\frac{\text{221}\text{.8378g}}{\text{245}\text{.8368g}}\\ \\ \text{product =37}\text{.78473 mg}\end{array}$

(d)

Interpretation Introduction

Interpretation:

• Reaction of ${\text{LaCoO}}_{\text{3+x}}$ with ${\text{H}}_{\text{2}}\text{O}$ to gives ${\text{La}}_{\text{2}}{\text{O}}_{\text{3}\left(\text{s}\right)}{\text{,Co}}_{\left(\text{s}\right)}{\text{and H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$. This reaction has to be balanced.
• Mass of solid product when $\text{41}\text{.8724}{\text{mg of LaCoO}}_3$ reacts completely has to be calculated.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 27.EE

The balanced equation is written as,

${\text{LaCoO}}_{\text{3}}{}_{\left(\text{s}\right)}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }\underset{\text{FM}\text{221}\text{.8378}}{\underbrace{\frac{\text{1}}{\text{2}}{\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+ Co}}_{\left(\text{s}\right)}}}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$

Mass of the product is found to be $\text{0}\text{.565}\text{g}$.

Explanation of Solution

To write: balanced equation of ${\text{LaCoO}}_{\text{3}}$ with ${\text{H}}_{\text{2}}\text{O}$

${\text{LaCoO}}_{\text{3}}{}_{\left(\text{s}\right)}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\to }\underset{\text{FM}\text{221}\text{.8378}}{\underbrace{\frac{\text{1}}{\text{2}}{\text{La}}_{\text{2}}{\text{O}}_{\text{3}}{}_{\left(\text{s}\right)}{\text{+ Co}}_{\left(\text{s}\right)}}}\text{+}\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}{\text{O}}_{\left(\text{g}\right)}$

To calculate: Mass of the product

$\begin{array}{l}\frac{\text{Final mass}}{\text{Initial mass}}\text{= }\frac{\text{Final mass}}{\text{41}\text{.8724}\text{mg}}\text{=}\frac{\text{221}\text{.8378g}}{\text{245}\text{.8369g+}\left(\text{15}\text{.9994}\right)\text{x}}\\ \\ \text{Final mass =}\frac{\left(\text{221}\text{.8378g}\right)\left(\text{41}\text{.8724}\text{mg}\right)}{\text{245}\text{.8369g +}\left(\text{15}\text{.9994}\right)\text{x}}\\ \\ \text{245}\text{.8369g + }\left(\text{15}\text{.9994}\right)\text{x =}\text{9288}\text{.881}\\ \\ \text{15}\text{.9994x =}\text{9288}\text{.881}\\ \\ \text{x =}\text{0}\text{.565g}\end{array}$

(e)

Interpretation Introduction

Interpretation:

In the formula of ${\text{LaCoO}}_{\text{3}}$, the value of x has to be calculated and formula of starting solid has to be written.

Answer to Problem 27.EE

The value of x is found to be $\text{0}\text{.0086}$.

Formula of starting solid is ${\text{LaCoO}}_{\text{3}}$.

Explanation of Solution

To calculate: value of x in the formula of ${\text{LaCoO}}_{\text{3}}$

Experimental final mass is equal to 37.7637 mg.  Substitute this value into mass equation derived in part (d), gives

$\begin{array}{l}\frac{\text{Final mass}}{\text{Initial mass}}\text{=37}\text{.7637mg =}\frac{\left(\text{221}\text{.8378g}\right)\left(\text{41}\text{.8724}\text{mg}\right)}{\text{245}\text{.8369g +}\left(\text{15}\text{.9994}\right)\text{x}}\\ \\ \text{x =}\text{0}\text{.008g}\end{array}$

(f)

Interpretation Introduction

Interpretation:

The product near $\text{500°C}$ has to be determined.

Answer to Problem 27.EE

The product nearby $\text{500°C}$ is found to be ${\text{LaCoO}}_{\text{2}\text{.5}}$.

Explanation of Solution

To found: The product nearby $\text{500°C}$ where the mass is approximately 40.51mg

Loss of mass at $\text{500°C}$ is approximately $\text{41}\text{.8724}\text{mg - 40}\text{.51mg =1}\text{.36mg}$ which is $\frac{1}{3}$ of the theoretical complete mass loss from ${\text{LaCoO}}_{\text{3}}$.

That is $\frac{\text{1}}{\text{3}}\text{of 4}\text{.0877 mg =1}\text{.362mg}$.

The plates at $\text{500°C}$ relates to the change of $\text{Cobalt}\left(\text{III}\right)\stackrel{}{\to }\text{Cobalt}\left(\text{II}\right)$ which is same as ${\text{LaCoO}}_{\text{3}}\stackrel{}{\to }{\text{LaCoO}}_{\text{2}\text{.5}}$.