Question

${\mathbf{LaCoO}}_{\mathbf{3}\mathbf{+}\mathbf{x}}$, has a perovskite structure (Box 19-1) with variable oxygen content. The figure shows the thermogravimetric curve observed when 41.8724m gwere heated in role="math" localid="1665035505591" $\mathbf{5}\mathbf{}\mathbf{vol}\mathbf{\%}\mathbf{}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{in}\mathbf{}\mathbf{Ar}\mathbf{.}\mathbf{La}\left(\mathrm{II}\right)$does not react, but cobalt is reduced to Co(s)

(a) What is the oxidation state of cobalt in the ideal formula ${\mathbf{LaCoO}}_{\mathbf{3}}$

(b) Write the reaction of role="math" localid="1665035588740" ${\mathbf{LaCoO}}_{\mathbf{3}}\mathbf{}\mathbf{with}\mathbf{}{\mathbf{H}}_{\mathbf{2}}$ to produce ${\mathbf{La}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{}\mathbf{with}\mathbf{}\left(s\right)\mathbf{,}\mathbf{Co}\left(s\right)\mathbf{,}\mathbf{and}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(s\right)$

(c) If $\mathbf{41}\mathbf{.}\mathbf{8724}\mathbf{mg}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{LaCoO}}_{\mathbf{3}}$ react completely, what will be the mass of product$\mathbf{(}{\mathbf{La}}_{\mathbf{2}}\mathbf{}{\mathbf{O}}_{\mathbf{3}}\mathbf{+}\mathbf{(}\mathbf{Co}\mathbf{)}$?

(d) Write the balanced reaction of role="math" localid="1665035782894" ${\mathbf{LaCoO}}_{\mathbf{3}\mathbf{+}\mathbf{x}}\mathbf{}\mathbf{with}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ to produce ${\mathbf{La}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}\mathbf{,}\mathbf{Co}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{.}\mathbf{}\mathbf{If}\mathbf{}\mathbf{41}\mathbf{.}\mathbf{8724}\mathbf{mg}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{LaCoO}}_{\mathbf{3}\mathbf{+}\mathbf{1}}$react completely; what will be the mass of solid product? Your answer will be an expression with x in it.

(e) From the observed product mass $\left(37.767\mathrm{mg}\right)\mathbf{}\mathbf{at}\mathbf{}\mathbf{700}\mathbf{°}\mathbf{c}$, find x in${\mathbf{LaCoO}}_{\mathbf{3}\mathbf{+}\mathbf{x}}$ Write the formula of the starting solid.

(f) In part (c), the mass lost from the ideal formula ${\mathbf{LaCoO}}_{\mathbf{3}\mathbf{+}\mathbf{x}}$, is 4.0877 mg. What is the product nearrole="math" localid="1665035980286" $\mathbf{500}\mathbf{°}\mathbf{C}$where the mass is -40.51 mg?

Short answer

The solutions for the given conditions are

(a)The oxidation state of $\mathrm{Co}\mathrm{in}{\mathrm{LaCoO}}_3$ is found to be +3.

(b) role="math" localid="1665036221823" ${\mathrm{LaCoO}}_{3\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_{2\left(\mathrm{g}\right)}\to \frac{1}{2}{\mathrm{La}}_2{\mathrm{O}}_{\underset{\underset{\mathrm{FM}221.8378}{\downarrow }}{3\left(\mathrm{s}\right)}}+{\mathrm{Co}}_{\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

(c)Mass of the product is found to be 37.78473mg

(d)The balanced equation is written as,

${\mathrm{LaCoO}}_{3\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_{2\left(\mathrm{g}\right)}\to \frac{1}{2}{\mathrm{La}}_2{\mathrm{O}}_{\underset{\underset{\mathrm{FM}221.8378}{\downarrow }}{3\left(\mathrm{s}\right)}}+{\mathrm{Co}}_{\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

Mass of the product is found to be 0.565g

Mass of the product is found to be 37.78473mg

(e)The value of x is found to be 0.0086.

Formula of starting solid is${\mathrm{LaCoO}}_3$

(f)The product nearby $500°\mathrm{C}$is found to be${\mathrm{LaCoO}}_{2.5}$

Step-by-step solution

Find the oxidation state of cobalt

(a)Given information:

Oxidation state of Oxygen is -2.

Oxidation state of Lanthanide is +3.

To find: Oxidation state of Cobalt in the formula of${\mathrm{LaCoO}}_3$

Let the oxidation state of cobalt be x.

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

Thus, oxidation state of is found to be +3.

Write the reaction of LzCoO3 with H2

(b)To write the balanced reaction of${\mathrm{LaCoO}}_3\mathrm{with}{\mathrm{H}}_2$

$\mathrm{LaCo}{\mathrm{O}}_3\mathrm{reacts}\mathrm{with}{\mathrm{H}}_2\mathrm{to}\mathrm{gives}{\mathrm{La}}_2{\mathrm{O}}_{3\left(\mathrm{s}\right)},{\mathrm{Co}}_{\left(\mathrm{s}\right)}\mathrm{and}{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

Thus reaction is,

${\mathrm{LaCoO}}_{3\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_{2\left(\mathrm{g}\right)}\to \frac{1}{2}{\mathrm{La}}_2{\mathrm{O}}_{\begin{array}{c}3\left(\mathrm{s}\right)\\ \downarrow \\ \mathrm{FM}221.8378\end{array}}+{\mathrm{Co}}_{\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{g}\right)}$

Calculate mass of the given product

(c)To calculate: mass of the product when $41.8724\mathrm{mg}\mathrm{of}{\mathrm{LaCoO}}_3$reacts completely.

From the balanced reaction it can be seen that,

$$\begin{gathered} 245.8369\mathrm{g}{\mathrm{ofLaCoO}}_3\mathrm{produces}221.8378\mathrm{gof}({\mathrm{La}}_2{\mathrm{O}}_3+\mathrm{CO}) \\ 41.8724{\mathrm{mgofLaCoO}}_3\mathrm{willproduce}\frac{221.8378}{245.8369}\times 41.8724\mathrm{mgof}({\mathrm{La}}_2{\mathrm{O}}_3+\mathrm{CO}) \\ 41.8724\mathrm{mgof}{\mathrm{LaCoO}}_3\mathrm{will}\mathrm{produce}37.7847\mathrm{mgof}({\mathrm{La}}_2{\mathrm{O}}_3+\mathrm{CO}) \end{gathered}$$

Calculate mass of the product

(d)$$\begin{gathered} {\mathrm{LaCoO}}_{3+\mathrm{x}\left(\mathrm{s}\right)}+\frac{3}{2}{\mathrm{H}}_{2\left(\mathrm{g}\right)}\to \frac{1}{2}{\mathrm{La}}_2{\mathrm{O}}_{\begin{array}{c}3\left(\mathrm{s}\right)\\ \downarrow \\ \mathrm{FM}221.8378\end{array}}+{\mathrm{Co}}_{\left(\mathrm{s}\right)}+\left(\frac{3}{2}+\mathrm{x}\right){\mathrm{H}}_2{\mathrm{O}}_{\mathrm{g}} \\ {\mathrm{MassofLaCoO}}_{\left(3+\mathrm{x}\right)}=245.8369+15.99\mathrm{xg} \end{gathered}$$

$$\begin{gathered} (245.8369+15.99\mathrm{x})\mathrm{g}{\mathrm{ofLaCoO}}_3\mathrm{produces}221.8378\mathrm{gof}({\mathrm{La}}_2{\mathrm{O}}_3+\mathrm{CO}) \\ 41.8724{\mathrm{mgofLaCoO}}_3\mathrm{willproduce}\frac{221.8378}{245.8369+15.99\mathrm{x}}\times 41.8724\mathrm{mgof}({\mathrm{La}}_2{\mathrm{O}}_3+\mathrm{CO}) \end{gathered}$$

Amount of product formedrole="math" localid="1665037162149" $=\left(\frac{221.8378\times 41.8724\mathrm{mg}}{245.8369+15.99\mathrm{x}\mathrm{g}}\right)$

 Step 5: Calculate the value of x in the formula of LaCoO3

(e)To calculate: value of x in the formula of${\mathrm{LaCoO}}_3$

Experimental final mass is equal to 37.7637mg.

Therefore,

$$\begin{gathered} 37.7637\mathrm{mg}=\left(\frac{221.8378\times 41.8724}{245\times 8369+15.99\mathrm{x}}\right) \\ \mathrm{x}=0.008 \end{gathered}$$

Find the product with mass value

(f)To find the product nearby ${\mathrm{LaCoO}}_3$where the mass is approximately 40.51mg

Loss of mass at $500°\mathrm{C}$is approximately $41.8724\mathrm{mg}-40.51\mathrm{mg}=1.36\mathrm{mg}\mathrm{which}\mathrm{is}\frac{1}{3}$of the theoretical complete mass loss from ${\mathrm{LaCoO}}_3$

That is $\frac{1}{3}\mathrm{of}4.0877\mathrm{mg}=1.362\mathrm{mg}$

The plates at $500°\mathrm{C}$relates to the change of $\mathrm{Cobalt}\left(\mathrm{III}\right)\to \mathrm{Cobalt}\left(\mathrm{II}\right)$which is same as ${\mathrm{LaCoO}}_3\to {\mathrm{LaCoO}}_{2.5}$