Question

Write the balanced chemical equation when Hydrochloric acid reacts with Sodium Bicarbonate?

Answer 1

Step1: Reacting to the given molecules:

• Hydrochloric acid reacts with Sodium Bicarbonate to form Sodium Chloride, Carbon Dioxide and water.
• The unbalanced chemical equation can be written as:
• $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{NaCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)} \\ \mathbf{Hydrochloric}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{} \\ \mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Bicarbonate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$

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: $\mathbf{2}$ Sodium atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$, $\mathbf{3}$ Oxygen atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$,$\mathbf{1}$ Carbon atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$, $\mathbf{1}$ Hydrogen atom from $\mathbf{HCl}$ and $\mathbf{1}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{2}$ Sodium atom from $\mathbf{NaCl}$, $\mathbf{2}$ 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 ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.

Step3: Balancing the reactant molecules by multiplying $\mathbf{2}$ by $\mathbf{HCl}$, $\mathbf{2}$ by $\mathbf{NaCl}$

• The chemical reaction is as follows:
• $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{2}\mathbf{}\mathbf{NaCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)} \\ \mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{} \\ \mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Bicarbonate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$
• On the reactant side: $\mathbf{2}$ Sodium atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$, $\mathbf{3}$ Oxygen atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$,$\mathbf{1}$ Carbon atom from ${\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}$, $\mathbf{2}$ Hydrogen atom from $\mathbf{HCl}$ and $\mathbf{2}$ Chlorine atom from $\mathbf{HCl}$
• On the product side: $\mathbf{2}$ Sodium atom from $\mathbf{NaCl}$, $\mathbf{2}$ 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 ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$.
• After multiplication, we have an equal number of molecules on both sides of the reaction.

Step4: Balanced chemical equation is as follows:

• The chemical reaction is as follows:
• $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathbf{HCl}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{Na}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{\to }\mathbf{}\mathbf{2}\mathbf{}\mathbf{NaCl}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{\left(}\mathbf{l}\mathbf{\right)} \\ \mathbf{Hydrochloric}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Sodium}\mathbf{} \\ \mathbf{}\mathbf{Acid}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Bicarbonate}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{Chloride} \end{gathered}$$

Hence, the equation is balanced.