Chapter 3, Problem 3.49QP

(a)

Interpretation Introduction

Interpretation: Different minerals of Iron-Silicate like Ferrosilite ${\text{FeSiO}}_{\text{3}}$, Fayalite ${\text{Fe}}_{\text{2}}{\text{SiO}}_4$ and greenalite ${\text{Fe}}_3{\text{Si}}_2{\text{O}}_5{\left(\text{OH}\right)}_4$ are given. The reactions showing the weathering of these minerals are to be balanced.

Concept introduction: According to the law of conservation of mass, it is not possible to either create or destroy the mass. Therefore, the mass and number of atom of each element must be balanced on both the sides of the reaction.

To determine: The balanced reaction for the given unbalanced reaction.

Answer to Problem 3.49QP

Solution

The balanced reaction for the given unbalanced reaction is shown below.

${\text{3FeSiO}}_{\text{3}}(s)+4{\text{H}}_2\text{O}(l)\to {\text{Fe}}_3{\text{Si}}_2{\text{O}}_5{\left(\text{OH}\right)}_4\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Explanation of Solution

Explanation

The steps involved in the balancing of the reaction are shown below.

The given unbalanced reaction is,

${\text{FeSiO}}_{\text{3}}(s)+{\text{H}}_2\text{O}(l)\to {\text{Fe}}_3{\text{Si}}_2{\text{O}}_5{\left(\text{OH}\right)}_4\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

The element that appears in only one reactant and product is chosen and is balanced first. In the given reaction $\text{Fe}$ appears in only one reactant and product. Therefore, $\text{Fe}$ is balanced by multiplying ${\text{FeSiO}}_{\text{3}}$ with a coefficient of $3$ at the left hand side of the reaction.

${\text{3FeSiO}}_{\text{3}}(s)+{\text{H}}_2\text{O}(l)\to {\text{Fe}}_3{\text{Si}}_2{\text{O}}_5{\left(\text{OH}\right)}_4\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

It is seen that Silicon also gets balanced on both the sides of the reaction. Therefore, directly Hydrogen is balanced multiplying water with a coefficient of $4$ at the left hand side as shown below.

${\text{3FeSiO}}_{\text{3}}(s)+4{\text{H}}_2\text{O}(l)\to {\text{Fe}}_3{\text{Si}}_2{\text{O}}_5{\left(\text{OH}\right)}_4\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

It is seen that Oxygen also gets balanced on both the sides of the reaction.

(b)

Interpretation Introduction

To determine: The balanced reaction for the given unbalanced reaction.

Answer to Problem 3.49QP

Solution

The balanced reaction for the given unbalanced reaction is shown below.

${\text{Fe}}_{\text{2}}{\text{SiO}}_4(s)+2{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}(l)\to 2{\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Explanation of Solution

Explanation

The steps involved in the balancing of the reaction are shown below.

The given unbalanced reaction is shown below.

${\text{Fe}}_{\text{2}}{\text{SiO}}_4(s)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to {\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

The element that appears in only one reactant and product is chosen and is balanced first. In the given reaction $\text{Fe}$, $\text{H}$ appears in only one reactant and product Therefore, $\text{Fe}$ is balanced by multiplying ${\text{FeCO}}_3$ with a coefficient of $2$ at the right hand side of the reaction.

${\text{Fe}}_{\text{2}}{\text{SiO}}_4(s)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to 2{\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

There is one Carbon atom at the left hand side while there are two Carbon atoms at the left hand side. Therefore ${\text{CO}}_2$ is multiplied with a coefficient of $2$ as shown below.

${\text{Fe}}_{\text{2}}{\text{SiO}}_4(s)+2{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to 2{\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

It is seen that Oxygen also gets unbalanced on both the sides of the reaction.

Hydrogen is balanced by multiplying water with a coefficient of $2$ as shown below.

${\text{Fe}}_{\text{2}}{\text{SiO}}_4(s)+2{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}(l)\to 2{\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

The Oxygen atoms also balanced on both the sides of the reaction.

(c)

Interpretation Introduction

To determine: The balanced reaction for the given unbalanced reaction.

Answer to Problem 3.49QP

Solution

The balanced reaction for the given unbalanced reaction is shown below.

${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+3{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}(l)\to 3{\text{FeCO}}_3\left(s\right)+2{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Explanation of Solution

Explanation

The steps involved in the balancing of the reaction are shown below.

The given unbalanced reaction is shown below.

${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to {\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

The number of atoms of each element is counted properly.

The element that appears in only one reactant and product is chosen and is balanced first. In the given reaction, $\text{Fe}$, $\text{Si}$ and $\text{C}$ appears in only one reactant and product. Therefore, $\text{Fe}$ is multiplied with a coefficient of $3$ at the right hand side of the reaction.

${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to 3{\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Now $\text{Si}$ is balanced by multiplying ${\text{H}}_4{\text{SiO}}_4$ with a coefficient of $2$.

${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to 3{\text{FeCO}}_3\left(s\right)+2{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Carbon is balanced by multiplying ${\text{CO}}_2$ water with a coefficient of $3$ as shown below.

${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+3{\text{CO}}_2\left(g\right)+{\text{H}}_2\text{O}(l)\to 3{\text{FeCO}}_3\left(s\right)+2{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Hydrogen is balanced by multiplying water at the left hand side with a coefficient of $2$.

${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+3{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}(l)\to 3{\text{FeCO}}_3\left(s\right)+2{\text{H}}_4{\text{SiO}}_4\left(aq\right)$

Conclusion

The balanced reactions for the given unbalanced reaction are shown below.

1. a. ${\text{3FeSiO}}_{\text{3}}(s)+4{\text{H}}_2\text{O}(l)\to {\text{Fe}}_3{\text{Si}}_2{\text{O}}_5{\left(\text{OH}\right)}_4\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$
2. b. ${\text{Fe}}_{\text{2}}{\text{SiO}}_4(s)+2{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}(l)\to 2{\text{FeCO}}_3\left(s\right)+{\text{H}}_4{\text{SiO}}_4\left(aq\right)$
3. c. ${\text{Fe}}_3{\text{Si}}_{\text{2}}{\text{O}}_5{(\text{OH})}_4(s)+3{\text{CO}}_2\left(g\right)+{\text{2H}}_2\text{O}(l)\to 3{\text{FeCO}}_3\left(s\right)+2{\text{H}}_4{\text{SiO}}_4\left(aq\right)$