Chapter 22, Problem 22.82QA

Interpretation Introduction

To write:

The balanced net ionic equation for the reaction of aluminum carbonate with hydrochloric acid.

Answer to Problem 22.82QA

Solution:

The net balanced ionic equation is

$A{l}_2(C{O}_3{)}_3\left(s\right)+6 H^{+}\left(aq\right) \to 2 A{l^{+3} \left(aq\right)}^{ }+3 H_{2}O \left(l\right)+3 C{O}_2 \left(g\right)$

Explanation of Solution

Aluminum carbonate is a base while hydrochloric acid, $HCl$ is an acid. When aluminum carbonate reacts with $HCl$, the salt aluminum chloride, and carbonic acid is formed. Carbonic acid is unstable and decomposes into water and carbon dioxide.  The reaction is written in molecular form as

$A{l}_2(C{O}_3{)}_3\left(s\right)+HCl (aq) \to A{lC{l}_3}^{ }\left(aq\right)+H_{2}O \left(l\right)+C{O}_2\left(g\right)$

Taking an inventory of the atoms on both the sides of the reaction,

$2Al+9O+1H+1 Cl +3C \to 1 Al+3O+2H+3Cl+1 C$

To balance the $Al$, we need to put a coefficient 2 to $AlC{l}_3$.

$A{l}_2(C{O}_3{)}_3\left(s\right)+HCl (aq) \to 2 A{lC{l}_3}^{ }\left(aq\right)+H_{2}O \left(l\right)+C{O}_2\left(g\right)$

Taking an inventory of the atoms on both the sides of the reaction,

$2Al+9O+1H+1 Cl +3C \to 2 Al+3O+2H+6Cl+1 C$

To balance the $Cl,$ we need to add a coefficient 6 to $HCl$.

$A{l}_2(C{O}_3{)}_3\left(s\right)+6HCl (aq) \to 2 A{lC{l}_3}^{ }\left(aq\right)+H_{2}O \left(l\right)+ C{O}_2\left(g\right)$

Taking an inventory of the atoms on both the sides of the reaction,

$2Al+9O+6H+6 Cl 3C\to 2 Al+7O+2H+6Cl+1C$

To balance the $H,$ we need to add a coefficient 3 to $H_{2}O$.

$A{l}_2(C{O}_3{)}_3\left(s\right)+6HCl (aq) \to 2 A{lC{l}_3}^{ }\left(aq\right)+{3 H}_{2}O \left(l\right)+C{O}_2\left(g\right)$

Taking an inventory of the atoms on both the sides of the reaction,

$2Al+9O+6H+6 Cl 3C\to 2 Al+7O+6H+6Cl+1C$

To balance the $C,$ we need to add a coefficient 3 to $C{O}_2$.

$A{l}_2(C{O}_3{)}_3\left(s\right)+6HCl (aq) \to 2 A{lC{l}_3}^{ }\left(aq\right)+{3 H}_{2}O \left(l\right)+3 C{O}_2\left(g\right)$

Taking an inventory of the atoms on both the sides of the reaction,

$2Al+9O+6H+6 Cl 3C\to 2 Al+9O+6H+6Cl+3C$

The reaction is now completely balanced.

The total ionic equation is written as

$A{l}_2(C{O}_3{)}_3\left(s\right)+6 H^{+}\left(aq\right)+6C{l}^{-}(aq) \to 2 A{l^{+3} \left(aq\right)+6 C{l}^{-}\left(aq\right)}^{ }+3 H_{2}O \left(l\right)+3 C{O}_2 \left(g\right)$

Cancelling out the spectator ions, the net ionic equation is

$A{l}_2(C{O}_3{)}_3\left(s\right)+6 H^{+}\left(aq\right) \to 2 A{l^{+3} \left(aq\right)}^{ }+3 H_{2}O \left(l\right)+3 C{O}_2 \left(g\right)$

This is the balanced net ionic equation for the neutralization reaction.

Conclusion:

The net ionic equation is obtained by cancelling out the spectator ions in the total ionic equation.