Chapter 26, Problem 26.33QP

Interpretation Introduction

Interpretation:

A balanced reaction for the given reactions has to be written.

Concept introduction:

• There is a law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
• First write the reaction from the given information.
• Then count the number of atoms of each element in reactants as well as products.
• Finally obtained values could place it as coefficients of reactants as well as products.

Answer to Problem 26.33QP

The balanced reaction for the given reactions is

(a) ${\text{2Na}}_{\text{(s)}}{\text{ + 2H2O}}_{\text{(l) }}\to {\text{2NaOH}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{}_{\text{(g)}}$

(b) ${\text{2NaOH}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(c) ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(s) }}{\text{+ 2HCl}}_{\text{(aq)}}\to {\text{2NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(d) ${\text{NaHCO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + HCl}}_{\text{(aq) }}\to {\text{NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(e) ${\text{2NaHCO}}_{\text{3(s)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s)}}{\text{ + CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

(f) ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s) }}\to \text{no reaction}$

Explanation of Solution

(a)

Sodium metal reacts with water to produce sodium hydroxide and hydrogen gas.

${\text{Na}}_{\text{(s)}}{\text{ + H2O}}_{\text{(l) }}\to {\text{NaOH}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{}_{\text{(g)}}$

Here, the reaction is unbalanced. So we need to balance it.

To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.

${\text{2Na}}_{\text{(s)}}{\text{ + 2H2O}}_{\text{(l) }}\to {\text{2NaOH}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{}_{\text{(g)}}$

(b)

An aqueous solution of $\text{NaOH}$ reacts with ${\text{CO}}_{\text{2}}$ to produce sodium carbonate and water.

${\text{NaOH}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Here, the reaction is unbalanced. So we need to balance it.

To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.

${\text{2NaOH}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(c)

Solid ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ reacts with $\text{HCl}$ solution to produce sodium chloride, carbon dioxide and water.

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(s) }}{\text{+ HCl}}_{\text{(aq)}}\to {\text{NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Here, the reaction is unbalanced. So we need to balance it.

To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(s) }}{\text{+ 2HCl}}_{\text{(aq)}}\to {\text{2NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(d)

Solid ${\text{NaHCO}}_{\text{3}}$ reacts with $\text{HCl}$ solution to produce sodium chloride, carbon dioxide and water.

${\text{NaHCO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + HCl}}_{\text{(aq) }}\to {\text{NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

To balance the reaction, calculate the number of atoms present in left side and right side. Finally, obtained values could place it as coefficients of reactants as well as products.  In this reaction equally conversion happened so that reaction is already balanced.

${\text{NaHCO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + HCl}}_{\text{(aq) }}\to {\text{NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(e)

Solid ${\text{NaHCO}}_{\text{3}}$ is heated to produce sodium carbonate, carbon dioxide and water.

${\text{NaHCO}}_{\text{3(s)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s)}}{\text{ + CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Here, the reaction is unbalanced.  So we need to balance it.

To balance the reaction, calculate the number of atoms present in left side and right side.  Finally, obtained values could place it as coefficients of reactants as well as products.

${\text{2NaHCO}}_{\text{3(s)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s)}}{\text{ + CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

(f)

At normal temperatures sodium carbonate won’t go decomposition.

${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s) }}\to \text{no reaction}$

When we heat the solid ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ at moderate temperature it won’t undergo decomposition and hence there is no reaction.

Conclusion

A balanced equation for the given reactions were written

(a) ${\text{2Na}}_{\text{(s)}}{\text{ + 2H2O}}_{\text{(l) }}\to {\text{2NaOH}}_{\text{(aq)}}{\text{ + H}}_{\text{2}}{}_{\text{(g)}}$

(b) ${\text{2NaOH}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(aq) }}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(c) ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}{}_{\text{(s) }}{\text{+ 2HCl}}_{\text{(aq)}}\to {\text{2NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(d) ${\text{NaHCO}}_{\text{3}}{}_{\text{(aq)}}{\text{ + HCl}}_{\text{(aq) }}\to {\text{NaCl}}_{\text{(aq)}}{\text{ + CO}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

(e) ${\text{2NaHCO}}_{\text{3(s)}}\to {\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s)}}{\text{ + CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

(f) ${\text{Na}}_{\text{2}}{\text{CO}}_{\text{3(s) }}\to \text{no reaction}$