Chapter 18, Problem 18.101QA

Interpretation Introduction

To:

Balance the given reaction and determine whether it is a redox reaction or not

Answer to Problem 18.101QA

Solution:

$2 KAlS{i}_3{O}_8\left(s\right)+{2 H}_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+4 Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

This reaction is not a redox reaction.

Explanation of Solution

1) Concept:

The reaction is balanced by taking an inventory of the atoms on both sides of the reaction. The $K, Al, C and S$ are balanced first by adding appropriate coefficients. The $O and H$ atomsare balanced at the end.

If the reaction involves oxidation and reduction of the species involved, then the reaction is a redox reaction. This can be determined by calculating the oxidation numbers of the elements on the left side and right side of the reaction.

2) Given:

The unbalanced reaction is

$KAlS{i}_3{O}_8\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

3) Calculations:

$KAlS{i}_3{O}_8\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

Taking an inventory of atoms, we get

$1K+1 Al+3 Si+11O+2H+1 C\to 2K+2 Al+3Si+14 O+4H+1 C$

To balance the $K and Al,$ we add a coefficient 2 in front of $KAlS{i}_3{O}_8\left(s\right)$

$2 KAlS{i}_3{O}_8\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

Taking an inventory of atoms, we get

$2K+2 Al+6 Si+19O+2H+1 C\to 2K+2 Al+3Si+14 O+4H+1 C$

To balance the $Si,$ we add a coefficient 4 in front of $Si{O}_2\left(s\right)$

$2 KAlS{i}_3{O}_8\left(s\right)+H_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+4 Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

Taking an inventory of atoms, we get

$2K+2 Al+6 Si+19O+2H+1 C\to 2K+2 Al+6Si+20 O+4H+1 C$

To balance the $H,$ we add a coefficient 2 in front of $H_{2}O \left(l\right)$

$2 KAlS{i}_3{O}_8\left(s\right)+{2 H}_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+4 Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

Taking an inventory of atoms, we get

$2K+2 Al+6 Si+20O+4H+1 C\to 2K+2 Al+6Si+20 O+4H+1 C$

The reaction is now completely balanced.

The final balanced reaction is

$2 KAlS{i}_3{O}_8\left(s\right)+{2 H}_{2}O\left(l\right)+C{O}_2\left(g\right)\to A{l}_2\left(S{i}_2{O}_5\right){\left(OH\right)}_4\left(s\right)+4 Si{O}_2\left(s\right)+K_{2}C{O}_3\left(aq\right)$

The oxidation numbers of the elements in the compounds are as follows:

Compound	Element	Oxidation number
$KAlS{i}_3{O}_8$	$Si$	$+4$
$Si{O}_2$	$Si$	$+4$
$C{O}_2$	$C$	$+4$
$K_{2}C{O}_3$	$C$	$+4$

It is seen that the oxidation number of the elements do not change.

Thus, this is not a redox reaction.

Conclusion:

The reaction is balanced by inspection and by taking an inventory of the atoms. The oxidation numbers of the elements do not change; hence, it is a not a redox reaction.