Chapter 12, Problem 12.95QP

(a)

Interpretation Introduction

To determine: A balance chemical equation for the conversion of the mineral anorthite to a mixture of three minerals.

Interpretation: A balance chemical equation for the conversion of the mineral anorthite to a mixture of three minerals is to be stated.

Concept introduction: Anorthite is one of the rarest members of the plagioclase series. The plagioclase is a useful mineral in the Earth’s crust.

Grossular, also called grossularite, is a calcium aluminium garnet that sometimes resembles the gooseberry.

Kyanite, also known as cyanite, is a silicate mineral that is formed during the regional metamorphism of clay-rich sediments.

Answer to Problem 12.95QP

Answer

The balanced chemical equation is,

$3{\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\left(s\right)\to {\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\left(s\right)+2{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\left(s\right)+{\text{SiO}}_2\left(s\right)$

Explanation of Solution

Explanation

The given three minerals are,

Grossular $\left[{\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\right]$,

Kyanite $\left[{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\right]$ and,

Quartz $\left[{\text{SiO}}_2\right]$

The transformation of given anorthite into a mixtures of three minerals is given as,

${\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\left(s\right)\to {\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\left(s\right)+{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\left(s\right)+{\text{SiO}}_2\left(s\right)$

According to the law of conservation of mass, the amount of reactants involved in a chemical reaction is equal to the amount of products formed in the chemical reaction.

The atoms of calcium, aluminium, silicon and oxygen are not equal on either side of the reaction. The coefficient $3$ is added before ${\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}$ and $2$ is added before ${\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}$ to balance the atoms present on either side of the stated reaction.

The balanced equation thus obtained is,

$3{\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\left(s\right)\to {\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\left(s\right)+2{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\left(s\right)+{\text{SiO}}_2\left(s\right)$

(b)

Interpretation Introduction

To determine: The charges and formulas of silicate anions in anorthite, Grossular, and Kyanite.

Interpretation: The charges and formulas of silicate anions in anorthite, Grossular, and Kyanite is to be stated.

Concept introduction: Anorthite is one of the rarest members of the plagioclase series. The plagioclase is a useful mineral in the Earth’s crust.

Grossular also called grossularite which is a calcium aluminium garnet that sometimes resembles the gooseberry.

Kyanite also known as cyanite it is a silicate mineral that is formed during the regional metamorphism of clay-rich sediments.

Answer to Problem 12.95QP

Answer

The charge on silicate anion in anorthite is ${\text{Si}}_{\text{2}}{\text{O}}_8{}^{8-}$.

The charge on silicate anion in grossular is ${\text{SiO}}_4{}^{4-}$.

The charge on silicate anion in kyanite is ${\text{SiO}}_5{}^{6-}$.

Explanation of Solution

Explanation

The given three minerals are,

Grossular $\left[{\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\right]$,

Kyanite $\left[{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\right]$ and,

Anorthite $\left[{\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\right]$

In anorthite $\left[{\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\right]$ the formula of silicate anion is ${\text{Si}}_{\text{2}}{\text{O}}_8$.

The oxidation state of calcium atom is $+2$.

The oxidation state of aluminium atom is $+3$.

The oxidation state of silicate anion in anorthite is assumed to be $x$.

The oxidation state of silicate anion in anorthite is calculated by the formula,

$\left(\begin{array}{l}\text{The sum of the oxidation states of }\\ \text{the constituent atoms in a complex}\end{array}\right)=\text{Charge on the complex}$

Substitute the given values in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{The sum of the oxidation states of }\\ \text{the constituent atoms in a complex}\end{array}\right)=\text{Charge on the complex}\\ \left(\begin{array}{c}\text{Oxidation}\text{state}\text{of}\text{Ca}+2\left(\text{Oxidation}\text{state}\text{of}\text{Al}\right)\\ +\text{Oxidation}\text{state}\text{of}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\end{array}\right)=0\\ 2+2\left(3\right)+x=0\\ x=-8\end{array}$

Therefore the charge on silicate anion in anorthite is ${\text{Si}}_{\text{2}}{\text{O}}_8{}^{8-}$.

In grossular $\left[{\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\right]$ the formula of silicate anion is ${\text{SiO}}_4$.

The oxidation state of calcium atom is $+2$.

The oxidation state of aluminium atom is $+3$.

The oxidation state of silicate anion in grossular is assumed to be $x$.

The oxidation state of silicate anion in grossular is calculated by the formula,

$\left(\begin{array}{l}\text{The sum of the oxidation states of }\\ \text{the constituent atoms in a complex}\end{array}\right)=\text{Charge on the complex}$

Substitute the given values in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{The sum of the oxidation states of }\\ \text{the constituent atoms in a complex}\end{array}\right)=\text{Charge on the complex}\\ \left(\begin{array}{c}3\left(\text{Oxidation}\text{state}\text{of}\text{Ca}\right)+2\left(\text{Oxidation}\text{state}\text{of}\text{Al}\right)\\ +3\left(\text{Oxidation}\text{state}\text{of}{\text{SiO}}_4\right)\end{array}\right)=0\\ 3\left(2\right)+2\left(3\right)+3x=0\\ x=-4\end{array}$

Therefore the charge on silicate anion in grossular is ${\text{SiO}}_4{}^{4-}$.

In kyanite $\left[{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\right]$ the formula of silicate anion is ${\text{SiO}}_5$.

The oxidation state of calcium atom is $+2$.

The oxidation state of aluminium atom is $+3$.

The oxidation state of silicate anion in anorthite is assumed to be $x$.

The oxidation state of silicate anion in anorthite is calculated by the formula,

$\left(\begin{array}{l}\text{The sum of the oxidation states of }\\ \text{the constituent atoms in a complex}\end{array}\right)=\text{Charge on the complex}$

Substitute the given values in the above expression.

$\begin{array}{c}\left(\begin{array}{l}\text{The sum of the oxidation states of }\\ \text{the constituent atoms in a complex}\end{array}\right)=\text{Charge on the complex}\\ 2\left(\text{Oxidation}\text{state}\text{of}\text{Al}\right)+\text{Oxidation}\text{state}\text{of}{\text{SiO}}_5=0\\ 2\left(3\right)+x=0\\ x=-6\end{array}$

Therefore the charge on silicate anion in kyanite is ${\text{SiO}}_5{}^{6-}$.

Conclusion

Conclusion

1. a. The balance equation is,

   $3{\text{CaAl}}_{\text{2}}{\text{Si}}_{\text{2}}{\text{O}}_{\text{8}}\left(s\right)\to {\text{Ca}}_3{\text{Al}}_2{\left({\text{SiO}}_{\text{4}}\right)}_3\left(s\right)+2{\text{Al}}_{\text{2}}{\text{SiO}}_{\text{5}}\left(s\right)+{\text{SiO}}_2\left(s\right)$.

2. b. The charge on silicate anion in anorthite is ${\text{Si}}_{\text{2}}{\text{O}}_8{}^{8-}$.

   The charge on silicate anion in grossular is ${\text{SiO}}_4{}^{4-}$.

   The charge on silicate anion in kyanite is ${\text{SiO}}_5{}^{6-}$.