Question

Write the balanced chemical equation for the following:

The action of Hydrochloric acid on Sodium bicarbonate.

Answer 1

Step1: Reacting the given molecules:

• When Hydrochloric acid is added to Metal carbonate/bicarbonate it will form their Metal chloride and water along with Carbon dioxide gas.
• The Chemical reaction for metal bicarbonate/carbonate with Hydrochloric acid is
• $$\begin{gathered} {\mathbf{NaHCO}}_{\mathbf{3}\mathbf{}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{NaCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{bicarbonate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{chloride}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

Step2: Balancing the chemical equation

• The total number of atoms of each element on the reactant side and the product side must be compared.
• On the reactant side: $\mathbf{1}$ Sodium atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{HCO}}_{\mathbf{3}}$, $\mathbf{3}$ Oxygen atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{HCO}}_{\mathbf{3}}$,$\mathbf{1}$ Carbon atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{HCO}}_{\mathbf{3}}$, $\mathbf{2}$ Hydrogen atom $\mathbf{1}$ from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{HCO}}_{\mathbf{3}}$ and $\mathbf{1}$ from $\mathbf{HCl}$ and $\mathbf{1}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{1}$ Sodium atom from $\mathbf{NaCl}$, $\mathbf{1}$ Chlorine atoms from $\mathbf{NaCl}$, $\mathbf{1}$ Carbon atom from ${\mathbf{CO}}_{\mathbf{2}}$, $\mathbf{2}$ Hydrogen atoms from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ and $\mathbf{3}$ Oxygen atoms from $\mathbf{2}$ Oxygen atom from ${\mathbf{CO}}_{\mathbf{2}}$ and $\mathbf{1}$ Oxygen atom from ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.

1. The equation is already balanced as we have an equal number of molecules on both sides of the reaction.

Step 3: Balanced chemical equation is as follows:

1. Balanced reaction of Hydrochloric acid with Sodium bicarbonate is

$$\begin{gathered} {\mathbf{NaHCO}}_{\mathbf{3}\mathbf{}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{NaCl}\mathbf{}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \\ \mathbf{bicarbonate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{chloride}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{} \end{gathered}$$